wli 



CAL DRAWING. 



' 



LIBRARY OF CONGRESS, 



©|ap, ©ojtttrigljt Ifa.- 

Stieif.3L&£3» 

UNITED STATES OF AMERICA. 



OCT 



3 181/ 



MECHANICAL DRAWING. 



PREPARED FOR THE USE OF THE STUDENTS 

OF THE 

MASSACHUSETTS INSTITUTE OF TECHN0L0GY ; 

BOSTON, MASS. 



LINUS FAUNCE. 






BOSTON : 
W. J. Schofield, Printer, 105 Summer Street. 

1887. 



Copyrighted, 1887, by Linus Faunce. 



'V 



CONTENTS. 



Page. 

CHAPTER I. — Instruments and their Uses 5 

II. — Geometrical Problems 13 

III. — Inking 55 

" Tinting 61 

IV. — Projections 65 

" Notation 68 

" Projections of Straight Lines 69 

" " " Surfaces 75 

" Solids 80 



V. — Shadows. 



96 



vi. — isometrical drawing 112 

" Oblique Projections 120 

VII. — Working Drawings 121 

VIII.— Examples 127 



MECHANICAL DRAWING. 



CHAPTER I. 

INSTRUMENTS AND THEIR USES. 

1. To do good work, good instruments are essential. An 
accomplished draftsman may do fair work with poor instruments, 
but the beginner will find it sufficiently hard to do creditable 
work without being handicapped by poor instruments. It is 
also essential that the instruments should be kept in good order. 
They should be handled carefully, and wiped, before being put 
away, with wash-leather or chamois skin. This is especially 
needful if the hands perspire perceptibly. 

2. Pencilling. Drawings should always be first made in 
pencil, and inked afterwards if desired. The idea of pencilling 
is to locate the lines exactly, and to make them of the required 
length. Accuracy in a drawing can only be obtained by accu- 
racy in the pencil construction. There is a great tendency 
among beginners to overlook this important fact, and to become 
careless in pencilling, thinking they will be able to correct their 
inaccuracies when inking. This is a great mistake, and one to 



6 INSTRUMENTS AND THEIR USES. 

be especially avoided. This accuracy can be obtained in pencil 
only by making very Jine, light lines, and to this end hard pen- 
cils, 6 H, should be used, and they should be kept sharp. For 
drawing straight lines the pencil should be sharpened to a 
flat, thin edge, like a wedge. The compass pencils should be 
sharpened to a point. A softer pencil, 4 H, sharpened to a 
point, should be used in making letters, figures, etc. 

It should be borne in mind that a 6 H pencil sharpened to a 
chisel point will make a depression in the paper, which can 
never be erased, if much pressure is put on the pencil ; hence, 
press very lightly when using a hard pencil, so as to avoid this 
difficulty. If a drawing is not to be inked, but made for rough 
use in the shop, or where accuracy of construction (in the draw- 
ing) is not essential, or to be traced, a soft pencil would prefer- 
ably be used, the lines being made somewhat thicker, or heavier. 

3. Compasses. In using the compasses the lower part of 
the legs should be kept nearly vertical, so that the needle point 
will make only a small hole in revolving, and both nibs of the 
pen may press equally on the paper. In pencilling it is not so 
essential that the pencil point be kept vertical, but it is well to 
learn to use them in one way, whether pencilling or inking. 

Hold the compasses loosely between the thumb and fore- 
finger only, and do not press the needle point into the paper. 
If it is sharp, as it should be, the weight of the compass will be 
sufficient to keep it in place. While revolving, lean the com- 
pass very slightly in the direction of revolution, and put a little 
pressure on the pencil or pen point. 

In removing the pencil or pen point to change them, be very 
careful to pull them out straight ; do not bend them from side to 
side, in order to get them out more easily, as it would enlarge 
the socket and consequently spoil the instrument for accurate 
work. 



INSTRUMENTS AND THEIR USES. 7 

In drawing a circle of larger radius than could be drawn with 
the compass in its usual form a lengthening bar is used. In this 
case steady the needle point with one hand and describe the 
circle with the other. 

The large compasses are too heavy and clumsy to make small 
circles nicely, hence the bow compasses should be used in mak- 
ing all circles smaller than three-quarters of an inch radius, or 
thereabouts, depending on the stiffness of the spring. Be very 
careful to adjust the needle point to the same length as the pen- 
cil or pen point. In changing the radius of the bow compasses 
or spacers, press the points together, thus removing the pressure 
from the nuts, before turning the nuts in either direction. The 
screw thread will last much longer if this is done. 

4. Dividers or Spacers. These are used to lay off distances 
from scales, or from other parts of a drawing to a line, or to 
divide a line into equal parts. In laying off the same distance 
several times on a line, keep one of the points of the dividers on 
the line all the time, and turn the instrument in an opposite 
direction each time, so that the moving point will pass alter- 
nately to the right and left of £n"e line. Bo not make holes in 
the paper in doing this, as it is impossible to ink nicely over 
them ; a very slight puncture is sufficient. 

5. T-Square. The T-square should be used with the head 
against the left-hand edge of the drawing board (unless the per- 
son is left-handed), and horizontal lines only should be drawn 
with it. Lines perpendicular to these should not be drawn by 
using the head of the T-square against an adjoining edge of the 
board, as there is no pains taken to make these edges at right 
angles to each other, but they should be drawn by using the 
triangle in connection with the T-square. 



8 



INSTRUMENTS AND THEIR USES. 



Lines should be drawn with the upper edge only of the 
T-square. 

In case you wish to use the T-square as a guide for the knife in 
cutting paper to size, do not use the upper edge as a guide, but 
turn the T-square over and use the bottom edge. For, unless 
you are very careful, the knife will nick the edge, which would 
render it unfit to draw lines with. 

6. Triangles. To draw lines which shall be parallel to 
another by means of the triangles. Let AB be the given line. 




Place either edge of either triangle so as to coincide exactly 
with the given line. Place the other triangle (or any straight 

edge) against one 'of the other edges of the first triangle. Then, 
holding the second triangle, or straight edge, securely in this 
position with the left hand, move the first one, still keeping the 
two edges in contact. Any line drawn along the edge which 
originally coincided with the line AB will be parallel to it. 

To draw lines which shall be perpendicular to another by means 
of the triangles. Let AK be the given line. Place the longest 
side of the triangle so as to coincide exactly with the given line. 



INSTRUMENTS AND THEIR USES. 9 

Place the other triangle (or any straight edge) against one of 
the other edges of the first triangle. Then, holding the second 
triangle securely in this position with the left hand, revolve the 
first one so that its third edge is against the second triangle or 
straight edge. Any line drawn along the edge which originally 
coincided with the given line AB will be perpendicular to that 
line. 

The right-hand portion of the figure shows how the two tri- 
angles may be used, in connection with the T-square, to draw 
lines making angles of 15° and 75° with a given line (in this 
case the line which coincides with the edge of the T-square). 
By turning the triangles over, these angles may be drawn in the 
opposite direction. 

Lines making angles of 30°, 45°, and 60° are drawn directly 
by means of one triangle and T-square or straight edge. 

7. Irregular Curves. To trace an irregular curve through 
a series of points, use that part of the edge of the curve which 
coincides with the greatest possible number of points (never less 
than three), and draw the curve through these points, then shift 
the curve so as to coincide with other points in the same wav, 
letting the instrument run back on a part of the curve already 
drawn, so that a continuous smooth curved line may be formed. 

It requires a considerable practice to draw irregular curves by 
means of an instrument, the tendency being to make a series of 
loops, on account of some of the points being covered up. 
There is no better way to put in a curve in pencil than by doing 
it free-hand, provided the hand and eye have been properlv 
trained. Of course the curve cannot be inked in free-hand ; the 
irregular curve must be used, but, being no longer confined to 
points, it is not difficult. 



10 INSTRUMENTS AND THEIR USES. 

8. Scales. As it is frequently impossible to make a draw- 
ing on paper the real size of the object, it is customary to reduce 
the actual measurements by means of an instrument called a 
scale, — that is, the drawing may be made -§-, J, -J, T l ff , etc., size, 
accordiug as the relative size of the object and drawing may 
require. 

If it is desired to make a drawing \ size, then 3 inches on 
the drawing will represent one foot on the object. It is fre- 
quently necessary to represent inches and fractions of an inch, 
hence divide the 3 inches into 12 equal parts, and each of these 
parts will represent one inch on the object. If each of the 12 
parts are subdivided into 2, 4, or 8 parts, each part would re- 
present respectively ^-, ^, or -J of an inch on the object. This 
may be designated, scale, 3 inches equal one foot, or \ size. 

On the scale, one inch equal one foot, the unit, one inch, is 
divided into 12 parts to represent inches as before. Thus, to make 
a scale of any unit to one foot, it is simply necessary to divide 
that unit into 12 parts to represent inches, subdividing these 
parts, as far as possible, to represent fractions of an inch. 

If the smallest division on a scale represents -J of an inch on 
the object, the scale is said to read to -J of an inch. 

The student will find on his triangular scale ten different 
scales, viz., ^-, -J, f\, £, f , ^, f , 1, 1^-, and 3 inches to the foot, 
reading to 3"" 1", 1", 1", 1", f , £", J", J", and £", respectively, 
(the double prime over a number or fraction means inches, the 
single prime indicates feet). 

The scale should never be used as a ruler to draw lines with. 

9. Needle Point. Each student should procure a fine 
needle, break off the eye end, and force the broken end into a 
small} round piece of soft pine wood. This is to be Used in 
pricking off measurements from the scale, marking the exact 



INSTRUMENTS AND THEIR USES. 



it 



Cap, . . . . 


13 x 17 inches. 


Demy, . . 


15x20 " 


Medium, . . 


17 x 22 " 


Royal, . . 


19x24 " 


Super-Royal, 


. 19 x 27 " 


Imperial, 


22x30 " 



intersection of two lines, etc. Here, as in the case of the needle 
point in the compasses, it should not be forced into the paper : 
the finest puncture possible is sufficient. 

10. Drawing Paper. This paper comes in sheets of 
standard sizes, as follows : — 

Elephant, . . 23 x 28 inches. 
Columbia, . . 23 x 34 
Atlas, . . . 26 x 34 
Double Elephant, 27 x 40 
Antiquarian, . 31 x 53 
Emperor, . . 48 x 68 

Whatman's paper is considered the best. This paper is either 
hot or cold pressed, the hot pressed being smooth and the cold 
pressed rough. The rough paper is better for tinting work, the 
smooth takes ink lines better than the rough, but erasures show 
much more distinctly on it, hence the cold pressed is better for 
general work. The names of the sizes of the paper given above 
have no reference to quality. There is very little difference 
in the two sides of the paper, but that one which shows the 
maker's name in water lines, when held up to the light, is con- 
sidered the right side. 

11. Thumb Tacks. These are used for fastening the paper 
to the drawing board when it is not necessary to stretch it. 

12. The geometrical problems in the next chapter are not 
given with the view of teaching geometry, but to give the stu- 
dent practice in the accurate use of his instruments. 

In order that the degree of accuracy of the execution of the 
problems may be readily seen, these problems will not be inked. 

13. The plates on which these problems are to be drawn 
should be laid out 10'' by 14" (and cut this size when finished) 



12 INSTRUMENTS AND THEIR USES. 

with a border line one inch from each edge. That portion of 
the plate within the border line is to be divided into 6 equal 
squares, in each of which one problem is to be drawn, beginning 
with No. 1 in the upper left-hand corner square, No. 2 in the 
upper middle, No. 3 in the upper right, etc. 

The number of the plate is to be printed in the upper right- 
hand corner of the plate, about one-eighth of an inch above the 
border line, and the student's name in the lower right-hand cor- 
ner, about one-eighth of an inch below the border line. 

Prob. 1, 11, etc., as the case may be, is to be printed in the 
upper right-hand corner of each square. The initial letters of 
the name should be made T 3 ^ of an inch high, and the small 
letters -§• of an inch high. 

The letters are to be made like the samples furnished in the 
drawing room, and they should be made as nicely as possible. 



CHAPTER IL 



GEOMETRICAL PROBLEMS. 



>'p 



Prob. 1 . To bisect a straight line AB, or arc of a circle AFB. 
With A and B as centres and any radius 
greater than one half AB draw arcs intersect- 
ing in C and D. Join CD. CD is perpen^ 
"B dicular to AB, and E and F are the middle 
points required. 

Note. To draw a perpendicular to a line 
at any point in it, as E in AB. Lay off equal distances EA 
and EB on each side of E, and proceed as above. 

Prob. 2. From a point C outside a straight line AB to draw 
a perpendicular to the line. 

With C as a centre and any convenient 
radius cut AB in the points A and B. With 
A and B as centres and any radius draw arcs 
intersecting in D. Join C and D, and CD is 
the perpendicular required. 



14 



GEOMETRICAL PROBLEMS. 



Prob. 3. To draw a perpendicular to a line AB from a point 
C nearly or quite over its end. 

\, Draw a line from C to meet AB in any 

point B. Bisect BC in D by Prob. 1. With 
D as a centre and radius DC draw the arc 
CAB meeting AB in A. Draw through A 
and C. 
A ^------ --' Note. If a perpendicular be required at 

A, take any point D as a centre and radius DA and draw an 
arc CAB. Through B and D draw a line to meet the arc in C. 
Draw through C and A. 





, 




—/{--■ 


/ 


4 




/ ' 


i 










A, 




^B_ 



Prob. 4. To draw a perpendicidar to a line AB from a point 
A at or near its end. 

With A as a centre and any radius draw 
the arc CD. With centre D. and same radius 
cut CD in C. With C as a centre and same 
radius draw an arc over A, and draw a line 
through D and C. producing it to meet this 
■ arc in E. Draw through A and E. 



■$' 



JL 



Prob. 5. A second method. 

With centre A and any radius draw an arc 
CDE ; with centre C and same radius cut this 
arc in D ; with centre D and same radius draw 
arc EF ; with centre E and same radius draw 
arc intersecting-. EF in F. Join AF. 



^TJN 



jQS £ 



GEOMETRICAL PROBLEMS. 



15 




Prob. 6. To construct an angle equal to a given angle CAB. 
Draw any line as DE, and take any point in 
it as D. With A as a centre and any radius 
cut the sides of the given angle in B and C. 
With centre D and same radius draw arc EF. 
With BC as a radius and E as a centre cut arc 
EF in F. Join DF. 

Note. For accuracy of construction in drawing this prob- 
lem the longer the radius AB is the better. * 

Prob. 7. Through a given point C to draw a line parallel to 
a given line AB. 
P^= .c_ From C as a centre and anjr radius 

\ y' \ draw the arc AD ; and from A as a centre, 

\ y' \ with the same radius, draw the arc BC. 

— ^~ ^- With BC as a radius, and A as a centre, 

draw an arc cutting the arc AD in D. Draw through D and C. 

Prob. 8. To draw a line parallel to a given line AB at a 
given distance CD from it. 

From any two points A and B on the line 
as centres and with CD as a radius draw arcs 
E and F. At A and B erect perpendiculars 
to meet the arcs in E and F. Draw a line 
thro u oh E and F. 



1 " ' ?jc~"^ 




/*"~pk 


A! 


:b 


C^_ 


n 



Prob. 9. To bisect a given angle BAC. 

With A as a centre and witli any radius 

draw the arc BC, cutting the sides of the angle 

>'■ in B and C. With centres B and C and any 

adius draw arcs intersectino; in D. Draw AD. 




16 



GEOMETRICAL PROBLEMS. 



Prob. 10. To divide a given line AB into any number of 
equal parts. (In this case 6.) 

,.>■* From A draw an indefinite line A, 1, 2 • • 5 

? . -f'f J\ at any angle with AB. At B draw Ba • ■ • e, 

A**/""/ : I / '. b making the angle ABe equal to the angle 

\ ; : i i-"' 6 - BA5. With any distance as a unit, lay off on 

V ' L 

r^l d - the lines from A and B as many equal spaces 

as the number of parts required less one. 

Join le, 2d, 3c, etc. 



Prob. 11. Another method. 
(say) Jive parts. 



To divide a line AB into 



Draw A 1 • • • • 5 at any angle to AB, and 
lay off on it five equal spaces, using any con- 
venient unit. Join 5B, and through the 
points 1, 2, 3, 4 draw lines parallel to B5, 
meeting AB in points a, b, c, d. 




Prob. 12. To divide a line AC into the same proportional 
parts as a given divided line AB. 

Draw from a point A the lines AC and 
AB, making any angle. Join B and C. 
Through the points 1, 2, and 3 on AB draw 
lines parallel to BC, meeting AC in points a, 
b, and c, the required points of division. 




GEOMETRICAL PROBLEMS. 



17 






Prob. 13. Second method. 

Let M be the line to be divided into parts 
proportional to the parts Al, 12, 23, and 3B 
of the line AB. Draw M parallel to AB at 
CD by Prob. 8. Draw lines through AC and 
BD to meet in E. Through E draw lines to 
1, 2, and 3, cutting CD in a, b, and c, the re- 
quired points of division. 

Note. If the parts on AB are equal, the parts on CD will 
be equal. 



r 


a! 


\h 




c 


n 




i 


> 






v \ 


i 




2 


M 


3 


B 



Prob. 14. To draiv on a given line AB as an hypothenuse a 
right triangle with its sides having the proportion of 3, 4, and 5. 
Divide AB by Prob. 11 into 5 equal parts. 
With centre B and a radius equal to three of 
the parts draw an arc, and with centre A and 
a radius equal to four of the parts cut this 
arc in C. Join AC and BC. 




Prob. 15. 




To trisect a right angle CAB. 

With centre A and any radius draw the arc 
of the quadrant, cutting the sides in C and B. 
With centres C and B and the same radius cut 
the arc in points 1 and 2. Join Al and A2. 

Note. The angle 1AB is an angle of 60°, 



the construction of: which is apparent. 



18 



GEOMETRICAL PROBLEMS. 



Prob. 16. To find a mean 'proportional between two lines M 
and N. 

Upon an indefinite line AB lay off AC equal 
to M, and CB equal to N. Bisect AB in E 
by Prob. 1, and with centre E and radius EB 
draw a semicircle. At C draw CD perpen- 
dicular to AB (Prob. 5). CD is the mean pro- 
portional required. 



-_n_, 



Prob. 17. To fend a fourth proportional to three lines M, N, 
and P. 

/£\ Draw AB equal to M, and at any conve- 

nient angle draw AC equal to N. Upon 

hrS. Ab__jcK AB produced make BD equal to P. Join 

— n BC, and draw DE parallel to BC, to meet 

AC produced in E. CE is the fourth pro- 
portional required. 



Prob. 18. 

30°. ' 




At a point A in a line AC to make an angle of 

With any point B as a centre and radius 
AB draw a semicircle ADC. With centre C 
and same radius cut this arc in D. Draw 
- AD. DAC = 30°. 



Prob. 19. Having given the sides AB, M, and N of a tri- 
angle to construct the figure. 

With centre A and radius M draw an arc. 
With B as a centre and radius N draw an arc 
to cut the first arc at C. Join AC and BC. 




GEOMETRICAL PROBLEMS. 



19 



Prob. 20. On a given side AB to construct a square. 



Draw BD at right angles to AB and equal 
to AB (Prob. 5). With A and D as centres 
and radius AB draw arcs intersecting in. C. 
Join AC and CD. 




Prob. 21. 
and M. 

£ d 



To construct a rectangle of given sides AB 

At B draw BD perpendicular to AB by 

Prob. 5, and equal to M. With A as a cen- 

■ T N tre and radius equal to M draw an arc, and 

from D as a centre and a radius equal to AB 

cut this arc in C. Join AC and CD. 



Prob. 22. On a given diagonal AB to construct a rhombus 
of given side AC. 

With centres A and B and radius AC draw 
arcs intersecting in C and D. Join AC, AD, 
b BC, and BD. 




Prob. 23. On a given base AB to construct a pentagon.* 1 ^ 
With centres A and B and radius AB 
draw circles intersecting in 1 and 2. Join 
1 and 2. With centre 2 and same radius 
draw the circle 3A5B4, giving points 
3 and 4. Produce 35 to C and 45 to 
E. With centres C and E and radius 
AB draw arcs intersecting in D. Draw 
BCDEA. 
jXote. This is an approximate method. 




20 



GEOMETRICAL PROBLEMS. 



Prob. 24. To construct a regular hexagon of given side AB. 

With A and B as centres and radius AB draw 
arcs intersecting in 0. With centre O and radius 
AB draw a circle, and lay off BC, CD, etc., 
each equal to AB. Join the points B, C, D, 
E, F, and A. 




Note. 
ence as a chord six times exactly. 



The radius of any circle goes around the circumfer- 




Prob. 25. On a given base AB to construct a regular poly- 
gon of any number of sides (in this case 7). 

With centre A and radius AB draw a 
semicircle and divide it at points 1, 2, 3, 4, 
etc., into as many equal parts as there are 
sides in the required polygon. Draw a line 
from the second point of division 2 to A. 

2 A is one side of the required polygon. 

Bisect AB and A2 by perpendiculars, by 

Prob. 1, meeting in D. With D as a centre and radius DA 

draw the circle BA2, etc. Apply AB as a chord to the circle as 

many times as there are sides in the polygon. 

-*«rROB. 26. On a given line AB to construct a polygon of any 
number of sides. An approximate method. 

Bisect AB, and produce the bisecting 
line indefinitely. With centre A and 
radius AB draw the arc BC, cutting the 
bisecting line in C. Divide the arc BC 
into six equal parts, in points 1, 2, 3, etc. 

To construct a pentagon. With cen- 
tre C and radius Cl draw an arc, cutting 
the bisecting line in a point below C, 




GEOMETRICAL PROBLEMS. 



21 



which is the centre of the circle circumscribing the pentagon. If 
the polygon is to have more sides than six, set up from C on the 
line Cab as many parts of the arc CB as added to six make 
the number of sides of the required polygon ; thus, for a seven- 
sided polygon set up one division as Ca ; for an eight-sided set 
up two divisions as Cab, and so on. a, b, c, d, etc., are the cen- 
tres for the circumscribing circles of the polygons, each side of 
which is equal to AB. 

On a given base AB to construct a regular octa- 

At A and B erect perpendiculars by 
Prob. 5 to AB, and bisect the exterior right 
angles, making the bisectors AC and BD 
;c each equal to AB. Draw CD, cutting the 
perpendiculars in E and F. Lay off EF 
from E to G and from F to H. Draw 
an indefinite line through GH. Make GK, GL, HN and HM 
each equal to CE or FD. Connect C, K, L, M, N, and D. 

Note. This may be done by Prob. 25. Check the con- 
struction by seeing if AN, CM, BD, and KL are parallel. 
CA, KB, etc. should be parallel ; so also AD, CN, and KM. 




-- IN 



Prob. 28. To construct an octagon within a square ABCD. 

Draw the diagonals AC and BD. With A, 
B, C, and D as centres and the half of the 
diagonal of the square as a radius draw the 
arcs EF, GH, KL, and MN. Join the points 
MG, FL, HN, and KE. 



D 

v" ,,K 


c 

5X 


E ~^X A 


r.-'-fi 


X 




£--''/' > 


> - - L 


/\ie 


Vs > . 



22 



GEOMETRICAL PROBLEMS. 



A A 


\ / 

\ // 


— ^N 


E. 


3 F 



Prob. 29. 77>e altitude AB o/" em equilateral triangle being 
given, to construct the triangle. 

Draw CD and EF, both perpendicular to 
AB. With A as a centre and any radius 
describe the semicircle CGHD. With C and 
D as centres and the same radius draw arcs 
cutting the semicircle in G and H. Draw 
AGE and AHF. 

Prob. 30. Given the base AB of an isosceles triangle, and 
the angle at the vertex M, to draw the triangle. 

With centre B and any radius draw a 
semicircle cutting the base produced at C. 
Make the angle DBC equal toMby Prob. 
6. Bisect ABD by Prob. 9. Make the 
angle FAB equal to EBA, and produce 
- AF and BE to meet in G. 




Prob. 31. 
a square. 




Within a regular hexagon ABCDEF to inscribe 



Draw a diagonal AD. Bisect AD by 
a perpendicular 213 (Prob. 1). Bisect by 
Prob. 9 the right angles 21A and 21B, and 
produce the bisectors to meet the sides of the 
hexagon in points G, H, L, and K. Join G, 
H, L, and K. 



Prob. 32. On a given diagonal AB to construct a square. 

Bisect AB in (Prob. 1) ; and with centre 
and radius O A draw a circle to cut the bisect- 
ing line in C and D. Draw AC, CB, BD, and 
DA. 




GEOMETRICAL PROBLEMS. 



23 



Prob. 33. Within a given triangle ABC to inscribe a square. 



G\ 



irk 



FG and GH. 



Draw AD perpendicular to and equal to AB. 
From C draw CE perpendicular to AB (Prob. 2). 
Draw DE cutting AC in F. From F draw FK 
perpendicular to AB. Make KH equal to KF, 
and from H with radius HK cut BC in G-. Join 



Prob. 34. About a given circle FEGD to circumscribe an 
equilateral triangle. 

Draw the diameter DE. With centre E 
and radius of the given circle draw the cir- 
cle AFG. Prolong DE to A. With centres 
D, F, and G and radius DG draw arcs inter- 
secting at B and C. Join AB and AC. 




Prob. 35. To circumscribe a circle about a triangle ABC. 

By Prob. 1 bisect two of the sides AB 
.j and BC by perpendiculars meeting in O. 
With centre O and radius OA draw the 
circle. 

Note 1. If any three points are given 
not in the same straight line, a circle is 
passed through them by joining the points 
and proceeding as above. 

Note 2. If any circle is given, its centre is found by taking 
any three points in its circumference, and proceeding as above. 




24 



GEOMETRICAL PROBLEMS. 




To inscribe a pentagon within a circle. 

Draw any diameter AB, and a radius CD 

perpendicular to it. Bisect BC in E. With 

centre E and radius ED draw the arc DF. 

B With centre D and radius DF draw the arc 

FG. DG is the side of the required pen- 



To inscribe a hexagon within a circle. 

Draw the diameter AB, and with centres A 
and B and the radius of the given circle 
draw arcs COD and EOF, cutting the circum- 
ference in C, D, E, and F. Join AD, DF, 
FB, etc. 

Note. Joining points A, E, and F gives 
an inscribed equilateral triangle. 



Prob. 38. To construct a regular polygon of any number of 
sides, the circumscribing circle being given. 

Draw any diameter AB and divide it 
into as many equal parts as there are sides 
in the required polygon (in this case 5). 
With A and B as centres and radius AB 
draw arcs intersecting in C. Draw a line 
from C through the second point of division 
of AB to meet the circumference in D. 
AD is one side of the required polygon. 
Lay off AD as a chord as many times as 
there are sides in the required polygon (in 
this case 5). 

Note. This is an approximate method. 




vfc'' 



GEOMETRICAL PROBLEMS. 



25 



To circumscribe a hexagon about a circle. 

Draw the diameters AB and CD perpen- 
dicular to each other. Divide each quadrant 
into thirds, by Prob. 15, at points E, F, etc. 
Join B and E, cutting CD in G. With G 
as a centre and radius GE draw arc ED, 
cutting COD in D. With centre O and 
radius OD cut the diameters produced in points H, K, C, L, and 
M. Join points H, D, M, L, C, and K. 

Note. Practically the 60° triangle placed on a T-square 
whose blade is parallel to COD will give HD, DM, etc., by 
making it tangent to the circle at E. F, etc. KH and LM are 
drawn by the T-square. 




Prob. 40. 
a circle. 




Prob. 41. 
without it. 



To circumscribe a square, also an octagon, about 

Draw the diameters AB and CD at rio-ht 
angles to each other. With centres A, B, C, 
and D and radius OA describe arcs intersect- 
ing in points E, F, G, and H. These points con- 
nected give a square about the circle. Inscribe 
an octagon in the square by Prob. 28. 

To draiv a tangent to a circle B from a point A 



Draw AB and bisect it in D. With 

centre D and radius DA draw a semicircle 

cutting the given circle in C and E. Join 

AC. By joining AE a second tangent 

is found, equal to AC. 

Note. To draw a tangent to a circle from a point C on the 

circumference. Join BC, and at C draw AC perpendicular to 

BC. For the tangent is perpendicular to the radius at the point 

of tangency. 




26 



GEOMETRICAL PROBLEMS. 



Prob. 42. To draw a tangent to the arc of a circle when 
the centre is not accessible. 

„ Let C be the point upon the given arc, 

AB, at which the tangent is to be drawn. 
Lay off equal distances upon the arc from 
C to A and B. Join A and B. Through C draw a line par-, 
allel to AB by Prob. 7. 




Prob. 43. To draw a tangent at a given point A on a circle 
when the preceding method is not applicable. 

From A draw any chord AB. Bisect AB in 
C, and the arc ADB in D by Prob. 1. With 
A as a centre and a radius AD draw the arc 
EF. With D as a centre and radius DF draw 
an arc cutting EF at E. Join AE. 

To draw a tangent to two given circles, A and B. 

Through A and B draw a line. 

Make Dfl equal to the radius BF. 

_T Draw the circle A-HC, and from B 

K ~ 

draw the tangent BC by Prob. 41. 
Draw AC and produce it to E. Make 
the angle FBK equal to CAH. Join EF. 



Prob. 44. 




Prob. 45. To draw a tangent to two given circles which shall 
pass between them. 

Join A and B, and draw AD and BC 
perpendicular to AB. Draw DC, cut- 
ting AB in E. Draw a tangent from 
E by Prob. 41 to the given circles. 
Join the tangent points F and G. FGr 
is the required tangent. 




GEOMETRICAL PROBLEMS. 



27 



Prob. 46. To draw a circle tangent to a given line AB at 
a given point B in it ,. which shall also pass through a fixed point 
C without the line. 

Draw BD perpendicular to AB, at the 
point B. Join CB, and draw a perpen- 
dicular to it at its middle point, Prob. 1. 
The intersection of this perpendicular and 
BD gives D the centre required. 





Prob. 47. To draw a circle of a given radius AB, which 
shall he tangent to a given circle C, and also to a straight line 
DE. 

Draw GH parallel to DE, by Prob. 
8, at the distance AB. With a radius 
CM, equal to the radius of circle C 
plus AB, draw an arc to meet GH in O. 
With the centre and radius AB draw 

£ § E the required circle. 

Note. If two circles are tangent, the straight line connect- 
ing the centres passes through the point of tangency. This 
point it is very important to locate precisely in all cases of tan- 
gency. 

Prob. 48. To draw a circle tangent to a given circle D, and 
also tangent to a given line AB, at a given point B on the line. 

Draw BG perpendicular to AB, and 

make BC equal to the radius of D. 

Join DC, and at its middle point draw 

a perpendicular to meet BC in E, the 

required centre. 

N ^c Note. By laying off BC above the 

line AB, and proceeding as above, another circle is found below 

AB. 




28 



GEOMETRICAL PROBLEMS. 



Prob. 49. To draw a circle tangent to a given circle A, and 
a given line BC, at a given point E on the circle. 

Draw AE and produce it. At E 
draw a perpendicular to AE, meeting 
BC in B. Draw the bisector of the 
angle EBC, meeting AE produced 
in D. DE is the radius required. 

Note. By bisecting EBM another 
circle, whose centre is G, is found, 
enclosing the circle AE. 




Prob. 50. To draw a circle of given radius R tangent to 
two given lines, AB and CD. 

Draw lines parallel to AB, and 
CD, at the given distance R from 
them by Prob. 8, meeting in E, 
the required centre. 



*' c 




Prob. 51. To draw amy number of circles tangent to each 
other, and also to two given lines AB and AD. 

Bisect BAD by AC, Prob. 9. Let 
one of the circles be BED, drawn by 
taking C as a centre, and a radius 
equal to the perpendicular CD from 
C to AD. At E draw EF perpendic- 
ular to AC. With centre F and ra- 
dius FE draw the arc DEG, cutting AD in G-. At G make 
GH perpendicular to AC. H is a centre required. Repeat 
the process. 




GEOMETRICAL PROBLEMS. 



29 



Prob. 52. To draw a circle through a given 'point D and 
tangent to two given lines AB and AC. 

Draw AD. Bisect BAC by AE. 
From any point K, on AE as a 
centre, draw a circle' tangent to AB 
and AC, and cutting AD in H. 
Draw HK. At D draw DE, mak- 
ing the angle ADE equal to AHK. 
E is the required centre. 




Prob. 53. To draw a circle of a given radius R tangent to 
two given circles A and B. 

Through A and B draw indefinite 
lines, and make DE and YQ each 
equal to R. With A and B as centres, 
and radii AE and BG, draw arcs cut- 
ting each other at C, the required cen- 
tre. 

R Note. These arcs will intersect in 

a second centre. 




Prob. 54. To draw a circle through a point C, and tangent 
to a given circle A, at a point B in its circumference.- 



Draw AB and produce it. Join BC 
,. and bisect it by a perpendicular meeting 
~---^ja» AB produced in D, the required centre. 




30 



GEOMETRICAL PROBLEMS. 



Peob. 55. Given two parallel straight lines AB and CD, to 
draw arcs of circles tangent to them at B and C, and passing 
through E, which is anywhere on the line BC. 

a At B and C erect perpendiculars. 
Bisect BE and CE by perpendiculars 
(Prob. 1), meeting the perpendiculars 
from B and C in F and H, the re- 
quired centres. Draw the arcs BE 

This is called a reversed curve. 




Prob. 56. To draw a circle tangent to two given circles A 
and B at a given point C in one of them. 

i*\ Draw a line indefinitely through 

A and C. Make CD and CE each 
equal to the radius of B. Join BE 
and BD. Bisect BE and BD by 
perpendiculars meeting AC pro- 
duced in F and G-, the centres of 
the required circles. 

Note. The tangent points should 
be determined accurately. 




Prob. 57. 




Same as Prob. 56. A second method. 

Draw a line through A and C 
indefinitely. Through B draw DE 
parallel to CA, cutting B in D and 
E.. Draw CD to F and CE, cut- 
ting B in H. Through B and H 
draw BHK, cutting CA in K, one 
centre required. Through B and F 
draw FBL, cutting CA in L, an- 



other centre required. 



GEOMETRICAL PROBLEMS. 



31 



Prob. 58. Same as Prod. 5Q. Prob. 59. Same as Prob. 56. 

Method of Prob. 57. 





Probs. GO and 61. Same as Prob. 56. Method of Prob. 57. 





Prob. 62. 




To inscribe a circle within a triangle ABC. 

Bisect two of the angles of the triangle, 

■^ say A and B, by Prob. 9. The bisectors 

meet in D, the centre of the required circle. 

A perpendicular from D to either side is the 

* required radius. 



32 



GEOMETRICAL PROBLEMS. 



Prob. 63. Within an equilateral triangle ABC to inscribe 
three equal circles, each touching two others, and two sides of the 
triangle. 

Draw the bisectors of the angles A, B, 
and C, cutting the sides in D, E, and F. 
With centres D, E, and F, and radius DF, 
draw arcs cutting the bisectors in H, L, 
and K, the required centres. 




Pkob. 64. Within an equilateral triangle to draw three equal 
circles, each touching two others and one side of the triangle. 

Bisect the angles A, B, and C. Bi- 
sect the angle DAB by AG. G is the 
centre of one of the required circles. 
With the centre O of the triangle as a 
centre, and radius OG, draw a circle 
cutting AD in H and BP2 in K. With 
centres G, H, and K, and radius GF, 
draw the circles G, H, and K. 




Pkob. 65. Within an equilateral triangle ABC to draw six 
equal circles which shall be tangent to each other and the sides of 

the triangle. 

Inscribe the three circles K, H, and 
G by Prob. 64. Draw LGM parallel 
to AB, MHN parallel to BC, and 
LKN parallel to AC. The points L, 
M, and N are the centres- of the other 
three circles. 




GEOMETRICAL PROBLEMS. 



33 



Prob. 66. Within a square ABCD to draw four equal cir- 
cles each touching two others and two sides of the square. 



Draw the diagonals AC and BD, and 
the diameters EF and GH. Draw EH, 
HF, FG, and GE, giving the centres 
M, N, O, and P. The radius OR is found 
by joining OP. 




Prob. 67. Within a given square ABCD to draw four equal 

circles each touching two others and one side of the square. 

$l ■- e c 

Draw the diagonals AC and BD, and 

the diameters EF and GH. Bisect the 
angle OAB by AK, cutting EF in K. 
H With radius OK and centre O draw a cir- 
cle cutting the diameters in the points L ? 
M, N, aud K, the required centres. 




Prob. 68. Within a square ABCD to draw four equal semi- 
circles, each tangent to two sides of the square, and their diame- 
ters forming a square. 

Draw the diagonals and diameters. Bi- 
sect FC and BH in L and K. Draw LK, 
cutting GH in M. Set off SM from S 
on the diameters to N, O, and P. Join 
the points M, N, 0, P. The intersections 
of these lines with the diagonals give the 
required centres 1, 2, 3, and 4. 




34 



GEOMETRICAL PROBLEMS. 



Prob. 69. Within a square ABCD to draw four equal semi- 
circles, each touching one side of the square and their diameters 
forming a square. 

4U: A Draw the diagonals and diameters. Draw 

EH, HF, FG, and GE, giving the points 
K, L, M, and N. The lines joining these 
points cut the diameters in points 1, 2, 3, 
and 4, the required centres. 



*"; 



r 



Prob. 70. To draw within a given circle ABC three equal 

circles tangent to each other and the given circle. 

Divide the circle injto six equal parts 
by diameters AE, DC, etc. (Prob. 24). 
Produce any diameter, as AE to G, mak- 
ing EG equal to the radius of the given 
circle. Join CG. Bisect the angle OGC 
by GH, intersecting OC in H. With 

centre O and radius OH draw the circle HKLM. K, L, and 

M are the centres of the circles. 




Prob. 71. To draw within a given circle ACDT1 four equal 
circles which shall be tangent to each other and the given circle. 

Draw the diameters AB and CD at right 
angles, and complete the square FBED. 
Draw the diagonal FE and bisect the angle 
FED by EG; FD and EG meet in G. 
With centre F and radius FG draw a cir- 
cle cutting the diameters in G, H, K, and L, 
Hr^a, the required centres. 




GEOMETRICAL PROBLEMS. 



35 




Prob. 72. Within a given circle AFD . . . . C to draw six 
equal circles tangent to each other and the given circles. 

Draw the diameters AB, CD, and EF, di- 
viding the circle into six equal parts (Prob. 
24, Note). Divide any radius as OB into 
three equal parts (Prob. 11), at points- F and 
G-. With- centre and radius OF draw a cir- 
cle giving the centres F, 1, 2, 3, 4, 5 required. 
A circle of the radius FB may be drawn 
from centre O tangent to the six circles. 
Note. The above is a special method. The general method, 
to draw any number of equal circles in a given circle, tangent 
to each other and the given circle, is to divide the circle by 
diameters into twice as many equal parts as circles required. 
From B, the extremity of any one of these diameters, draw 
a tangent ST, Prob. 41, Note. Produce the diameters on 
each side of AB to meet the tangent in S and T. Bisect the 
angle T. The bisector meets OB in F, the centre of one of the 
required circles. Or F may be obtained by making TM equal 
to TB, and at M drawing a perpendicular to OT, meeting OB 
in F. With centre O and radius OF draw a circle cutting the 
diameters in points 1, 2, 3, 4, etc., the centres required. 



Prob. 73. About a given circle to circumscribe any number 
of equal circles tangent to each other and the given circle. 

Divide the circumference of the given 
circle by diameters into twice as many parts 
as circles required. From the extremity B 

| of any diameter draw a tangent SBT (Prob. 

41, Note), and produce the diameters on 
each side of OB to meet SBT in S and T. 
Produce OT, making TN equal to TB. 
Make NR perpendicular to TN, meeting 




3b GEOMETRICAL PROBLEMS. 

OB produced in R, the centre of one of the required circles. 
The other centres are at the intersection of a circumference 
drawn with radius OR and centre O, and every other diameter 
produced. 



Prob. 74. Within a given circle AC . . . E to draw any num- 
ber of equal semicircles touching the given circle, and their diam- 
eters forming a regular polygon. 

In the given circle let OA and OB 
be two radii at right angles. Divide 
the given circumference, commencing 
at B, into twice as many parts as semi- 
circles required, and draw diameters to 
the points of division. Join BA. BA 
cuts the first diameter to one side of OB 
at G. G is one end of two adjacent 
diameters required. Lay off OG from O on every other diam- 
eter in points H, K, etc. Join HK, KG, etc. 
the centres of the required semicircles. 




1, 2, 3, etc., are 



Prob. 75. To divide a circle into any number of parts which 
shall be equal in area. 

Let the number of parts be four. Divide 
a diameter into twice as many equal parts 
as areas required, in this case eight, by 
H points 1, 2, 3, 4, etc. With 1 and 7 as cen- 
tres and radius 01 describe a semicircle on 
opposite sides of the diameter ; with cen- 
tres 2 and 6 and radius 02 do the same 
thing, and so continue taking each point as a centre and the dis- 
tance from it to the end of the diameter as a radius. 




GEOMETRICAL PROBLEMS. 



37 



Prob. 76. To divide a circle into concentric rings having 

equal areas. 

Divide the radius CD into as many 
equal parts as areas required (Prob. 11) in 
points 1, 2, 3, etc. On CD as a diameter 
draw a semicircle, and at the points. 1, 2, 
3, etc. draw lines perpendicular to CD, 
meeting the semicircle in points A, B, etc. 
With centre C and radii CA, CB, etc. 

draw concentric circles. 




Prob. 77. A chord AB and a point C being given, to find 
other points in the arc of the circle passing through A, B, and C 
without using the centre. 

Draw AB, AC, and BC. Suppose 
four more points are required. With 
any radius and centres A and B draw 
the arcs DE and FG. Make CAE 
equal CBA, and GBC equal CAD. 
Divide the arcs DE and FG into the 
same number of equal parts, one 
more than the number of points required.. Number the points 
1, 2, 3, 4, from D toward E, and from G toward F. Draw 
lines from A and B through these points, and those passing 
through like-numbered points meet in points on the arc ACB. 
To construct a point on the curve below AB lay off GO equal 
to Gl on the arc FG, and D9 equal to Dl. Lines through 
and B, and A and 9 meet in K, a point on the curve. 




38 



GEOMETRICAL PROBLEMS. 




Prob. 78. On a chord AB to construct the supplementary 
arc to ACB, ivithout using the centre, C being a point on the arc. 

Join AC, AB, and BC. 
Make BAD and ABF each' 
equal to ACB. With A and 
B as centres, and any radius, 
draw arcs HD and KF. Di- 
^ vide the arcs DH and KF 
\ into the same number of equal 
parts (say five) by the points 
1, 2, 3, etc. Number the 
points as shown. Draw lines from A through 1, 2, 3, etc. to 
meet lines through the same numbers drawn from B. The lines 
through like numbers meet in points on the required arc. 

Prob. 79. To construct any number of tangential arcs of 
circles, having a given diameter. 

Suppose three (four) arcs are re- 
quired. Upon the given diameter 
as a base draw an equilateral triangle 
ABC (square ABCD). With each 
vertex as a centre, and a radius of 
half a side, draw arcs of circles tau^ent to each other, as shown. 




Prob. 80. At a point C on a line AB to draw two arcs of 
circles tangent to AB, and to two parallels AD and BE, form- 
ing an arch. 

Make AD equal to AC, and BE equal to BC. 
At C make CGr perpendicular to AB, and at D and 
E draw the perpendiculars DF and GE, meeting 
CG in F and Gr, the required centres. 




GEOMETRICAL PROBLEMS. 



39 



Prob. 81. To draw any number of equidistant parallel 
straight lines between two given parallels, AB and CD. 

Draw any line Ao at any angle to AB, 
and lay off on it any convenient dis- 
tance Al as a unit as many times as 
lines required, plus one. Thus, if' four 
parallels are required, lay off Al five 
times. With A as a centre and Ao as 
radius draw an arc cutting CD in point e. Join Ae. With 
centre A and radii Al, A2, A3, etc., cut Ae in a, b, c, etc. 
Through a, b, c, etc. draw parallels to AB and CD. 




Prob. 82. To draw through a point C a line to meet the 
inaccessible intersection of two lines, AB and DE. 

~ — ^ ^~ r F From C draw any lines CA 

% E ~^~ and CD to the given lines. 

Join AD, AC, and CD. Make 
BE parallel to AD, BF par- 
allel to AC, and EF parallel 
intersection of EF and BF gives a point F in 




to CD. The 
the required line 



Draw through C and F. 



Prob. 83. To draw a perpendicular to a line AB, which 
shall pass through the inaccessible intersection of two lines, AE 



Produce AB to cut the lines in A and C. 
From A draw a perpendicular to CD, and 
from C a perpendicular to AE ; these per- 
pendiculars meet in 0. Draw BO perpen- 
dicular to AB. OB passes through the 
intersection of CD and AE. 




40 



GEOMETRICAL PROBLEMS. 



Prob. 84. To draw an involute of a square ABCD. 

Produce the sides as shown. With 
centre C and radius CD draw the arc DE. 
With centre B and radius BE draw the arc 
EF. With centre A and radius AF draw 
the arc FG, etc. 

Note. Suppose a line to be wrapped 
around and in the direction of the peri- 
meter of any plane figure. Let the line be unwound, keeping 
it always straight in the process of unwinding. Any point in 
the line describes an involute. The involute of polygons is 
composed of arcs of circles, as in Prob. 84. 




Prob. 85 To draw the involute of a circle. 

Divide the circumference into any num- 
ber of equal parts, as at A, B, C, D, etc., 
and draw radii to these points. At A, B, 
C, D, etc. draw tangents. Let the curve 
start at A. On the tangent at B lay off a 
distance from B to 1 equal to one of the 
parts into which the circumference is di- 
On the tangent at C lay off a distance equal to two 
parts to 2. On the tangent at D three parts to 3 ; from E four 
parts to 4, etc. The curve through these points, 1, 2, 3, 4, etc., 
is the involute of the circle. 




vided. 



GEOMETRICAL PROBLEMS. 



41 



Prob. 86. To draw a spiral composed of semicircles, the 
radii being in arithmetical progression. 

Draw an indefinite line, BAC. On the line 
take any two points, A and B, as centres. 
,~ With A as a centre and radius AB draw a 
semicircle. With B as a centre and radius BC 
draw a semicircle, and so on, using A and B as 
centres, and taking the radii to the end of the 
diameter of the last-drawn semicircle. 




Prob. 87. To draw a spiral composed of semicircles, whose 
radii shall he in geometrical progression. 

Let the ratio be 2. Let AB be the radius of 
the first circle, A its centre. Draw the semicir- 
cle BLC. With B as a centre and radius BC 
draw the semicircle CMD. With C as a cen- 
m tre and radius CD draw the semicircle DXE. 

D is the next centre, the diameter DE of the last-drawn circle 
becoming the radius for the next circle. So proceed. 




Prob. 88. To draw a spired of one turn in a given circle. 

Divide the circle into any num- 
ber of equal parts, say twelve, by 
the lines OA, OB, OC, etc., and a 
radius OA into the same number 
of equal parts by the points 1, 2, 
3, 4, etc. With O as a centre 
and radius 01 draw an arc cutting 
OB in M ; with centre O and radius 
02 an arc cutting OC in N, radius 
Oo an arc cutting OD in point P, 
etc. Through M, N, P, etc. draw the curve. 
Note. This is " the spiral of Archimedes." 




42 



GEOMETRICAL PROBLEMS, 



Pkob. 89. To draw in a given circle a spiral of any num- 
ber of turns, say two. 

Draw radii dividing the circle 
into any number of equal parts. 
Divide any radius, as OA, into as 
many equal parts as turns in the 
spiral, and divide each part into 
as many equal portions in points 
1, 2, 3, 4, etc. as the circle is di- 
vided into. With centre and 
01, 02, 03, etc. as radii, draw 
arcs to meet the radii OB, OC, 
OD, etc. in points of the required curve. 





Prob. 90. Given the axes AB and CD of an ellipse to draw 
the curve, and at any point on the curve to draw a tangent. 

Place the axes AB and CD 
at right angles to and bisecting 
< each other at O. With centre 
C and radius OA cut AB in 
F and F', which are the foci. 
Between and F', or F, take 
any point G, dividing AB into 
two parts. With centres F and 
F' and radius AG draw arcs on either side of AB. With the 
same centres and radius BG draw arcs intersecting those drawn 
with radius AG, at points L, M, N, and P, which are points on 
the curve. Take any other point, T on AB, and repeat the 
above operation ; and so on until as many points as are neces- 
sary are found. Through the points draw the curve. FM 
plus F'M equal AB. Let M be the point at which the tangent 
is required. Produce FM and draw the bisector of the angle 
SMF'. MR is the tangent required. 



GEOMETRICAL PROBLEMS. 



43 



Note. For drawing the ellipse and similar curves through 
a series of points the so-called French Curves are to he used. 

Note. The major axis is sometimes called the transverse, 
and the minor the conjugate, axis. 



Prob. 91. To draw an ellipse by means of a trammel, the 
axes being given. 

Let the semi-axes be OA and OB. Mark 
off on the straight edge of a slip of paper 
or card MP equal to OA, and NP equal to 
,OB. Keep the trammel with the point N 
always on the major axis, and the point M 
on the minor axis, and P will be a point in 
many points as necessary, and draw the curve. 




the curve. Find 



Prob. 92. To draiv an ellipse, having given the axes. 

Let the semi-axes be OA and OB. With 
radii OA and OB and centre draw circles. 
Draw any radii. OM, ON, etc. Make MP, 
NT, etc. perpendicular to OA, and HP, KT, 
etc. parallel to OA. P, T, etc. are points 
on the curve. 




Prob. 93. 




To draw an ellipse, having given the axes. 
_j) Place the axes at right angles at their 
centres, and on them construct a rectangle, 
i* one half being shown in BDEC. Divide 
OA and DA into the same number of 
equal parts by points 1 , 2, 3, etc. Draw 
ines through C and 1, 2, 3, etc. to meet 
lines from B drawn to 1, 2, 3, etc. on AD. 
P, Q, R, etc. are points on the curve. 



44 



GEOMETRICAL PROBLEMS . 



Prob. 94. To draw a curve approximating to an ellipse. 

Draw the squares ABDC and BEFD, and 
their diagonals, intersecting in G and H. 
With centres G and H and radius GA draw 
the arcs AC and EF. With centres B and 
D and radius DA draw the arcs CF and AE. 




Prob. 95. To draw on a given line, AB as a major axis, a 
curve approximating an ellipse. 

Divide AB into three equal parts (Prob. 
11) by points C and D. With centres C and 
D and radius CA draw two circles intersect- 
ing in E and F. Through C and D draw 
EGG, EDII, FCK, and FDL, meeting the 
circles in points G, H, K, and L. With centres E and F and 
radius EG draw the arcs GH and KL, completing the curve. 




Prob. 96. Given the major and minor axes of an ellipse, to 
draw the curve approximately. 

Let CA be the semi-major axis, and BC the 
semi-minor axis. Join A and B. Make CD 
equal to BC, and BE equal to AD. Bisect 
AE by a perpendicular, meeting BC pro- 
duced in F. With centre F and radius FB 
draw the arc BH, and with centre G and 
radius GH draw the arc HA. 
Note. One quarter of the whole curve is only shown, leav- 
ing to the student the construction of the full ellipse. 




GEOMETRICAL PROBLEMS. 



45 




V*> 



Prob. 97. Having the axes given, to draw a curve of tan- 
gential arcs of circles approximating to the ellipse. 

AO is tbe semi-major axis, and OB 
the semi-minor axis. Draw a rectangle 
with the axes as sides. AOBC is one- 
quarter of the rectangle. Draw AB. 
From C draw CMP perpendicular to 
AB, and meeting BO produced in P. 
Make OE equal to OB. On AE as a 
♦ P diameter draw a semicircle AKE. Pro- 

duce OB to K. Make OL equal to BK. With centre P and 
radius PL draw the arc LN. Make AD equal to OK, and 
with centre M and radius MD draw the arc DN, meeting LN 
in N. Draw MR and PNS. With centre M and radius MA 
draw AR ; with centre N and radius NR draw RS, and with 
centre P and radius PS draw an arc from S through B» Repeat 
in each of the quadrants. 



Prob. 98. To draw a parabola when the abscissa AB and 

the ordinate BC are given. 

Draw the rectangle ABCD, and divide 
AD and DC into the same number of 
equal parts. Through the points of divis- 
ion on AP draw parallels to AB, and 
from A draw lines to the points on DC. 
The first line above AB meets the line 
from A to the first point of division from 

D in a point P on the curve. The second parallel to A meets 

the second from A to DC and so on. P, Q, R, and C are points 

in the curve. Repeat the same below AB. 




46 



GEOMETRICAL PROBLEMS. 



Prob. 99. To draw a 'parabola when the directrix AC and 
the focus D are given, and to draiv a tangent at any point L on 
the curve. 

This curve is such that its apex E 
is always half way between A and D, 
and the distance from D to any point 
upon the curve, as F, is always equal 
to the horizontal distance from F to 
the directrix. Thus DF equals FG, 
and DH equals HK, etc. Through 
D draw BDA perpendicular to AC. 
This is the axis of the curve. Draw parallels to AC through 
any points in AB, and with centre D and radii equal to the hori- 
zontal distances of these parallels from AC cut the correspond- 
ing verticals, which will give points on the curve. 

To draw the tangent at L. Draw the ordinate LN, meeting 
AB in N. Produce BA to the right. Make ET equal to EN. 
Draw LT, the tangent required. 




Prob. 100. To draw an hyperbola when the diameter AB, 
the abscissa BC, and the double ordinate DE are given. 

Complete the rectangle 
BCDF, and divide CD and 
DF each into the same num- 
ber of equal parts. Draw 
BL, BM, and BN, inter- 
secting lines AK, AH, and 
AG respectively in points 
on the curve. Repeat below 
and in the other half of the 
curve as indicated. 





GEOMETRICAL PROBLEMS. 



47 




-^fesAD 



To draw an oval on the diameter of a given 

Let AB bo the diameter. Draw the circle 
ACB. Make OC perpendicular to AB. Draw 
the lines BCD and ACE indefinitely. With 
centres A and B and AB as a radius draw the 
arcs BE1 and AD. With centre C and radius 
CD draw the arc DE. 

Note. The centres A and B may be taken anywhere on 
the line AOB produced. 

Prob. 102. Upon a given line AB to draw an oval. 

Bisect AB at C, and draw 
the perpendicular CD. With 
B as a centre and radius AB 
describe an arc AD. Bisect 
the quadrant AE in F. 
Through F draw BFG. AG 
is the first part of the curve.. 
Bisect CB in H, and draw 
HD. K is the second centre. 
Bisect EL in M, and draw KMN. GN drawn from Iv is the 
second part of the curve. Bisect CH in O, and draw DO. P is 
the third centre. From P through E draw PET. NT is the 
third part of the curve. From E with a radius ET carry the 
curve to the line DC, and repeat the operation for the other 
half of the curve. Draw a semicircle on the diameter AB for 
the other part of the oval. 

The cycloid is the path described by any point in the circum- 
ference of a circle which rolls along a straight line. 

An epicycloid is the path described by any point in the cir- 
cumference of a circle which rolls along the outside of another 
circle. 




48 



GEOMETRICAL PROBLEMS. 



A hypocycloid is the path described by any point in the cir- 
cumference of a circle which rolls along the inside of another 
circle. 

The rolling circle is called the generatrix, or generating cir- 
cle, and the line (straight or curved) on which it rolls is called 
the directrix. 

Every point in the tire of a wheel which rolls along the 
ground in a straight line describes a cycloid. 

Hence it is easily seen that in one revolution or turning 
around of the circle, or wheel, the circumference will roll out 
into a straight line. To lay off the circumference on the straight 
line, either calculate its length or divide the circle into equal 
arcs, and lay off on the straight line as many divisions as there 
are in the circle ; each division on the straight line being equal 
to the length of one division on the arc. 

The arc is more than the chord, so a distance greater than 
the chord must be taken, and this is obtained by judgment or 
by approximation. By dividing the arc into a number of smaller 
arcs, so that the chord practically coincides with the arc, the 
small chord being laid off as many times as there are small arcs, 
a length of straight line is obtained very nearly equal to the 
arc given. 



Prob. 103. To construct a cycloid. 




Let AB be the directrix, and AD 
the generating circle. Divide the 
rolling circle into any number of 
equal parts, say 12, and lay off 
these lengths of ares along AB, 
giving points a, b, c, etc. Through 
1, the centre of the generating cir- 
cle, draw a line parallel to AB. 
This is the line of centres. On 



GEOMETRICAL PROBLEMS. 



49 



this lay off 12, 23, 34, etc., equal to A«, oft, etc., and with 
centres 2, 3, etc. draw the generating circle in all its posi- 
tions tangent to AB at points a, ft, e, etc. Draw through the 
points of division on the rolling circle parallels to AB, to meet 
the different positions of the rolling circle in points P, R, S, T 
and U. These parallels are drawn in the figure from points 
p, r, s, U and u. Repeat the process for the other half of the 
curve. 

Another method is to take the chords of the arcs Bp, Br Bs 
etc., and with centres a, b, c, etc. cut the respective circle's in 
points P, R, S, etc. The chord aP equals Bp ; bR equals Br • 
cS equals Bs, etc. 

Note. When the number of divisions of the rolling circle 
is large the curve may be drawn by arcs of circles by taking a 
as a centre, and radius aA, and drawing from A to P. Pro- 
duce Pa and Rb to meet, giving the centre for arc PR - Rft ail(i 
Sc meet at the centre of arc Rs, etc. 

To construct an exterior epicycloid. 

Let R-A be the rolling circle on 
the outer circumference of the direct- 
ing circle. Divide R-A into any 
number of equal parts (say 12), and 
lay off these parts on Aab, etc., 
giving points a, b, c, d, etc. 

With the centre of the directing 
circle as a centre, draw an arc from 
R giving the line of centres R123 
etc. Draw from the centre of the 
directing circle radial lines through 
«i o, c, d, etc., meeting the line of centres in points 1, 2, 3, etc. 
the centres of the different positions of the rolling circle.' With 




50 GEOMETRICAL PROBLEMS. 

centres 1, 2, 8, etc. and radius RA draw the several positions 
of the rolling circle. With the centre of the directing circle 
as a centre, draw arcs through the points of division of the cir- 
cle R-A to meet the several. positions of the rolling circle in 
points C, D, E, F, etc., which are points on the curve. Draw 
through C, D, E, F, etc. 

The points of the curve may be obtained by drawing from a 
as a centre and radius equal to the chord of one division of R— A 
an arc to meet the second position of the rolling circle in C ; 
from b with radius equal to the chord of two divisions an arc 
to meet the third position in D ; from c with radius equal to 
the chord of three divisions to meet the fourth position in E, 
etc. Or from a lay off on the rolling circle tangent at a one 
part of R-A (in this case T \>-) ; on the one tangent at b two 
parts of R-A ; on the one tangent at c three parts, etc. 



Prob. 105. To construct a hypocycloid. 

See the curve on the interior of the directing circle in figure 
with Prob. 104. Aa, ab, be, etc. are equal parts of the circum- 
ference of the rolling circle. R123 etc. is the line of centres. 
The process and directions are the same as for the epicycloid. 

Note. When the diameter of the rolling circle is equal to 
the radius of the directing circle the hypocycloid becomes a 
straight line. 



GEOMETRICAL PROBLEMS. 



51 



Prob. 106. To construct an interior epicycloid. 

Let the circle A-C roll on B-CMN. 
On the circumference of B-C lay off 
equal arcs CM, MN, NO, etc. Draw 
from M, N, 0, etc. radial lines 
through B, and make MD, NE, OF 
each equal to the radius CA. With 
centres D, E, F, etc. draw the cir- 
cles tangent at M, N, O, etc. Lay 
off Ml equal to CM, giving point 1 ; 
from N lay off two divisions each 
equal to MC, giving point 2 ; from 

V Jay off three spaces, giving point 3, etc. 1, 2, 3, etc. are 

points of the curve. 




Prob. 107. To draw a scroll for a stair-railing. 

The circle ABCD is the eye of the scroll. 
Draw the diameters AC and BD at right 
angles. Draw the chord AD, and bisect 
it in 6. Draw a line 65 parallel to AC. 
Bisect AE in F. Bisect EF in 3. Make 
uj E4 equal to E3. Draw 32 and 45 paral- 
lel to BD, and 21 parallel to AC. From 
6 draw an indefinite line parallel to BD 
and produce 65, 21, etc. From point 1 
with radius IB draw the arc BH. From 
2 and radius 2H draw HJ, and so on. The arc BR of the inner 
curve is drawn from P with radius PB; the arc RS is drawn 
from 6 with radius 6R. 




52 



GEOMETRICAL PROBLEMS. 



Prob. 108. Another method. 

Suppose the scroll to come between 
the outside of the rail AB and the line 
CD. Draw A9 perpendicular to AB 
and CD, and divide it into nine equal 
parts. On the sixth division as a side 
draw the square abcQ. With centre a 
and radius aA. draw the quadrant AE. 
With centre 6 and radius 6E draw arc 
EF. With centre c and radius cF draw 
FG ; with centre 6 and radius 6G draw arc GH ; with centre e 
and radius eH draw arc HK ; and with centre /and radius fK 
draw the arc KL. For the inner curve start with a as a centre 
and radius al and proceed as for the outer. 




Prob. 109. 
given. 



To construct a spiral, its greatest diameter AB 




Divide AB into eight equal parts 
by points 1,2, 3, etc. On 45 as a 
diameter draw a circle CDEF. This 
circle is the eye of the spiral. In- 
scribe a square CDEF, as shown in 
the enlarged drawing of the eye. Draw 
the central diameters of the square, 
12,10 and 11,9. Divide these diam- 
eters into six equal parts, and number 
as shown. With 1 as a centre and ra- 
dius 1H draw an arc HK to meet a horizontal produced from 
2 through 1. With 2 as a centre and radius 2K draw the arc KL 
to meet a vertical through 3 and 2. With centre 3 and radius 3L 
draw arc LM meeting a horizontal produced through 4 and 3. 
With 4 as a centre and radius 4M draw arc MN to meet a line 
drawn through 5 and 4 at N, and so proceed. 



GEOMETRICAL PROBLEMS. 



53 



The curve may be commenced by taking 12 as a centre and 
radius 12 A and drawing from without toward the centre. 



Prob. 110. To describe an Ionic volute. 

Let AB be the vertical 
measure of the volute. Di- 
vide AB into seven equal 
parts, and from C, the lower 
extremity of the fourth di- 
vision, draw CF perpendic- 
ular to AB, of indefinite 
length. From any point on 
CF as a centre, with a ra- 
dius equal to one-half of 
one of the divisions of AB, 
draw the circle HIJK, form- 
ing the eye of the volute. Draw the diameter HJ perpendicu- 
lar to CF. Draw the square HIJK, bisect its sides, and draw 
the square 12 L M 11. Draw the dividing lines of the square 
as shown in the smaller figure, and extend them. The divis- 
ions corresponding to 12 N are equal. The divisions OP and 
RL are each equal to one-half of 12 N. From 1 as a centre 
and radius 1H draw the arc HS ; from 2 as a centre and radius 
2S the arc ST ; from centre 3 and radius 3T the arc TU ; and 
so proceed in the order of the numerals. 

In drawing the inner curves the dots on the diagonals in the 
small figure indicate the centres. The division of the square, 
of which 12 N is one side, shows how these centres are found. 




54 



GEOMETRICAL PROBLEMS. 



Fillet 



C Bead 




N \ Cavetto 




Torus 





Scotia 





Cyma Recta 



Cyma Reversa 



The Roman moldings are given above, the method of con- 
struction being evident. All the arcs are arcs of circles, and 
the angles are 45°, except in the flatter form of the cyma recta, 
where the line of centres is at 30°. 



CHAPTER III. 



INKING. 



14. It is supposed that the student has now become familiar 
with the use of the instruments necessary for the construction 
of a drawing in pencil, and has acquired a certain degree of 
proficiency in handling them which is necessary for accurate 
work. The next step is to learn to ink a drawing after it has 
been pencilled. 

Starting with a good pen, in good condition, and a smooth, 
well-ground black ink. it only remains for the student to learn 
to make a clean, sharp, even line. This may seem at first like 
an easy thing to do ; nevertheless, the ability to make a good 
ink line every time comes to most students very slowly, and 
after a great deal of practice. Therefore, before beginning to 
work on the plates, which are to be finished carefully and 
handed in for inspection, it will save a considerable time and 
paper to make lines against the triangle without regard to their 
length, direction, or location, until the student is thoroughly 
familiar with his pen and can make a fair line. Several hours, 
if necessary, in this preliminary practice will be invaluable. 

15. India Ink. A special ink, called India ink, is always 
used in making drawings. It comes either in the stick form, 
and has to be ground as used, or in the liquid form. In the 



56 INKING. 

latter form it is held in solution by. an acid which corrodes the 
pen and eats into the fibre of the paper, so that if it is desired 
to erase a line it is much more difficult than if made with the 
ground ink. This ink is also very liable to rub off like soot. 
The only advantage it has is the saving- of time in preparation. 
This kind of ink cannot be recommended for anything except 
coarse, rough work. It should not be used for tinting. 

To prepare the stick ink for use place a small amount of 
water in the ink slab or saucer (the slab should be perfectly 
clean), then grasp the ink firmly and, with a rotary motion, 
grind until the liquid is Mack and a little sluggish in its motion. 
After it has reached the point when it is black enough the 
grinding should cease, as a continuation only makes the liquid 
thicker, thus causing it to flow less freely from the pen. The 
liquid will look black in the slab after a very little grinding, but 
the necessary consistency will not be reached for some time. 
In order to determine when the proper point is reached make 
a heavy line with the drawing pen on a piece of paper and wait 
for it to dry ; do not go over this line a second time. If the 
ink has not been ground sufficiently, it will look pale after it is 
dry, in which case more grinding is necessary. If at any time 
the ink becomes too thick, it can be diluted by putting in more 
water and mixing thoroughly. The stick should always be 
wiped dry after using, to prevent its crumbling. The ground 
ink should be kept covered as much as possible to prevent evap- 
oration, which would soon cause it to become too thick for use. 
It is not advisable to prepare a large quantity of ink at once, 
as the greater the amount the longer it takes, and freshly 
ground ink is preferable. If carefully covered, however, it may 
be kept two or three days. In case the ink becomes dry in the 
saucer it should all be washed out, as it is almost impossible to 
redissolve it entirely so that there will not be little scales which 
get into the pen and cause the ink to flow irregularly. 



INKING. 57 

16. Drawing Pen. This instrument, commonly called a 
right-line pen, is one of the most important of the drawing 
instruments, and it is very essential that it be of good quality. 
The screw is used to adjust the distance between the nibs, in 
order to make the line of the desired weight. The ink may 
be placed between the nibs by means of a brush or strip of 
paper, but it is more convenient to dip the pen into the ink, 
being careful to wipe the outside of the nibs before using. 

While inking the pen should be held so that both nibs rest 
on the paper evenly, and it should be inclined a little to the 
right, or in the direction of the line, with its flatter side against 
the triangle or straight edge, the end of the middle finger rest- 
ing on the head of the screw. A slight downward pressure is 
necessary (the greater the rougher the paper), but do not press 
against the ruler, as the lines would be uneven in thickness. 
The ruler is simply a guide for the pen. The lines should 
always be drawn from left to right (relative to the person and 
not to the drawing). If it is desirable to go over a line a sec- 
ond time for any reason, it should be drawn in the same direc- 
tion ; never go backward over it. 

In inking a curved line by means of the right-line pen and 
irregular curve, it is necessary to constantly change the direc- 
tion of the pen so that the nibs shall always be tangent to the 
curve. This requires considerable practice to do nicely. 

In case the ink does not flow freely from the pen, moisten the 
end of the finger and touch it to the end of the pen, and try it 
on a piece of waste paper. If this fails the pen should be 
wiped out cleau and fresh ink put in. In making fine lines the 
nibs of the pen are near to each other, consequently the ink 
dries between them quite rapidly, hence it will be found advis- 
able to clean out the pen thoroughly quite frequently to insure 
perfect lines. This is one of the secrets of being able to make 



58 INKING. 

good fine lines ; they should also be made more rapidly than 
heavy lines ; the heavier the line the slower the pen should be 
moved. Do not keep the point of the pen too near the straight 
edge, as the ink is liable to flow against it, thus causing a blot. 
Especial attention should be given to the care of the pens - 
they should always be carefully wiped after using, and should 
not be put away with any ink dried on them, nor allowed to 
get rusty on the inside of the nibs. Any old piece of cotton 
cloth will answer to wipe the pens and stick of ink on. 

17. How to Sharpen the Pen. To make good lines 
the pen must be kept in first-class condition, — that is, not only 
clean, but sharp ; and every draftsman should be able to 
sharpen his own pen. 

The curve at the point of the nibs of the pen should always 
be a semi-ellipse, with its long diameter coinciding with the axis 
of the pen ; it should not be a semicircle, nor should it be 
pointed. The student is advised to look carefully at the points 
of his new pen, so as to get a correct idea of the proper curve 
before it becomes changed by wear. When this curve becomes 
changed by wear, or if, from any other cause, one nib is longer 
than the other, the nibs should be screwed together, then, hold- 
ing the pen in a plane perpendicular to the oil-stone, draw it 
back and forth over the stone, changing the slope of the pen 
from downward and to the right to downward and to the left, 
or vice versa, for each forward or backward movement of the 
pen, so as to grind the points to the proper curve, making them 
also of exactly the same length. • 

This process, of course, makes the points even duller than 
before, but it is a necessary step. Next separate the points a 
little by means of the screw, and then place either blade upon 
the stone, keeping the pen at an angle of about 15° with the 



INKING. 59 

face of the stone, move it backward and forward, at the same 
time giving it an oscillating motion, until the points are sharp. 
This is quite a delicate operation, and great care should be exer- 
cised at tirst. The pressure upon the stone should not be very 
great, and it is well to examine the point very often so as to be 
sure and stop when each nib has been brought to a perfect 
edge, otherwise one nib is liable to be longer than the other 
and the pen will not work well, even if each nib is sharp of 
itself. 

Although the points want to be brought to a perfect edge, 
they should not be sharp like a knife, as in that case they would 
cut the paper. It will probably be necessary to try the pen 
with ink to be sure that it is in good condition. Sometimes a 
slight burr is formed on the inside of the blades ; this is re- 
moved by separating the points still farther, so as to insert the 
knife-edge of the oil-stone between them and draw it carefully 
through. One motion should be sufficient to remove the burr. 

The pen should never be sharpened by grinding the inside 
of the blades other than just indicated. 

18. Inking a Drawing. In inking a drawing it is prefer- 
able to ink all the circles and arcs first, as it is easier- to make 
the straight lines meet the arcs than the reverse. Of a number 
of concentric circles the smallest should be inked first. Here, 
as in the case of the pencil compasses, the pen point should be 
kept nearly vertical, the top of the compass being inclined a 
little toward the direction of revolution, and there should be a 
slight downward pressure on the pen point, but none on the 
needle point. 

Where a large number of lines meet at a point care should 
be taken to avoid a blot at their intersection. These lines 
should be drawn from rather than toward the point, and each 
line should be thoroughly dry before another is drawn. 



60 INKING. 

Id case of two lines meeting at a point neither line should 
stop before reaching the point, nor go beyond it. Either of 
these defects gives a very ragged appearance to the drawing. 

19. Stretching Paper. For ordinary small line drawings 
it is usually sufficient to fasten the paper to the board by meaijs 
of thumb tacks, but for large drawings, or those which are to 
be tinted at all, it is necessary to stretch the paper by wetting 
it and fasten it to the board with mucilage. To do this lay 
the paper on the board, fold over about one-half an inch along 
each edge of the sheet ; do not cut the corners ; next wet the 
upper surface of the paper, except that portion folded over ; do 
not rub the surface of the paper, simply press the sponge against 
it on all parts ; apply the mucilage to one of the edges and 
fasten that edge down, beginning at the middle and rubbing 
toward either end ; do the same with the opposite edge next, 
giving a slight pull to the paper as it is fastened down ; repeat 
this process for the two remaining edges. 

It is very important that the edges of the paper where the 
mucilage is to be applied should be kept dry, so that the muci- 
lage will be ready to act as soon as it can dry, and to facilitate 
this the less mucilage you can use and accomplish the result 
the better. If a large quantity of mucilage is used it will 
moisten the edges of the paper so much that it will be likely 
not to stick, as the body of the paper will dry as soon as the 
edge, and therefore pull them up. The drying of the mucilage 
can be hastened by rubbing the edge briskly with a piece of 
thick paper under the fingers until it becomes hot. The board 
should never be placed near the fire or radiator to hasten the 
drying, as it would dry the paper before the mucilage set, caus- 
ing the edges to be pulled up. The board should be left to 
dry in a horizontal position, and all the superfluous water should 



TINTING. 61 

be removed with a sponge, so as to avoid water marks in the 
paper, which always show in tinted drawings. In some cases, 
when the mucilage sets slowly, it may be necessary to moisten 
the centre of the paper sometime after stretching, to prevent 
its pulling up the edges by drying too rapidly. 

20. Correcting and Cleaning Drawings. Pencil lines 
are removed by means of a piece of rubber. When a mistake 
is made in inking, or it is desired to change a completed draw- 
ing, it becomes necessary to erase an ink line. This can be done 
by means of a rubber ink eraser, the same as in the case of 
pencil lines, except that much more rubbing is necessary. Ink 
lines can also be removed, and more quickly, by means of a 
knife ; in this case care should be taken not to use the point of 
the knife, as V-shaped holes are made which will always show. 
The flat portion of the knife should be used. After erasing an 
ink line, the surface which has been made rough by scratching 
should be rubbed down with some hard, perfectly clean, rounded 
instrument before inking other lines over it. 

A drawing can be cleaned by means of India rubber, or stale 
bread crumbled on the drawing and rubbed over it. Although 
dirt can be removed from a drawing, it should be the aim & of 
the draftsman to keep it as clean as possible. Therefore, the 
drawing should be kept covered when not being worked upon, 
and, if the drawing is a large one, all except that portion which 
is in use should be kept covered. 

TINTING. 

21. Tinting may be done in colors or India ink, as desired. 
The method of putting on the tint is the same in either case; 
consequently, we will take up only the India-ink tint here. 

If a drawing is to be tinted, the paper must be stretched as 



62 TINTING. 

explained in the first part of this chapter. Especial care 
should be taken to keep the paper perfectly clean. That por- 
tion of the drawing which is to be tinted must not be touched 
with the India rubber, as the surface is thereby made rough 
and will not take a uniform tint. Hence, in laying out the 
work pencil lines must not be made on the surface to be tinted. 

22. Preparation of the Tint. Clean the ink slab, water 
glass, and the brushes thoroughly, also be sure that there are 
no scales on the stick of ink which could possibly come off. 
Fill the slab about half full of water, and grind the ink as pre- 
viously explained until it is black, but not thick. Fill the water 
glass about half full of clean water, and with a brush transfer 
enough of the ink in the slab to the glass to make a light tint. 
It is hard to get an ink which is absolutely free from specks, 
therefore, it is well to let the ink, after it is prepared in the 
slab, stand a short time to allow these specks to settle to the 
bottom ; then, in transferring the ink to the glass, do not plunge 
the brush down to the bottom of the slab, thus taking up this 
sediment, but let the brush fill from the surface of the liquid. 

The mixture in the water glass is the one to be used for 
tinting, and it is better not to make it as dark as you wish it 
upon the drawing when finished, as it is much easier to put on 
a light tint evenly than a dark one. The required depth of 
shade can be obtained by successive washes. Let each wash dry 
thoroughly before putting on another. A smoother effect can 
usually be obtained, especially on a large surface, by going over 
the surface to be tinted with clean water first, and then letting 
it dry. 

23. Laying on the Tint. Having laid out the surfaces 
to be tinted, incline the board so as to slope like an ordinary 



TINTING. 63 

desk ; then dip your brush into the tint you have mixed, and 
take up as much of the liquid as it will carry, begin in the upper 
left-hand corner of the surface and draw it along the upper 
boundary, holding the brush nearly vertical and leaving quite a 
puddle as you proceed. Lead this puddle gradually downward 
by going across the surface from left to right, about a quarter 
of an inch at a time, dipping the brush frequently in the tint 
so as to keep the puddle about the same size all the time ; it 
should not be large enough to run down at any point. This 
puddle should not be left standing at any place any longer than 
is absolutely necessary, as it is very apt to leave a streak ; 
therefore, having commenced on a surface, finish it as quickly as 
possible, — do not let anything interrupt you. 

When you get to the bottom dry your brush on a piece of 
blotting paper, and with the dry brush take up the superfluous 
tint, as you would with a sponge, until it looks even. 

In laying on the tint do not bear on with the brush, as the 
brush marks would be liable to show, but use the point only, 
just touching it to the paper so as to wet it, and the tint will 
follow along of itself (the board being properly inclined). 

In following the boundaries of the surface to be tinted let 
the brush be pointed towards the boundary from the inside of 
the surface. 

This gives what is called & flat tint, and is used to represent 
surfaces which are parallel to the plane of projection. 

24. For representing surfaces which are oblique to the plane 
of projection a graduated tint is necessary. There are two 
methods of doing this, — the French and the American. 

The French method consists in dividing the surface into small 
divisions (these divisions should be indicated only, not drawn 
across the surface), laying a flat tint on the first space, and 
when this is dry laying another flat tint on the first two spaces. 



64 TINTING. 

Proceed in this way until the whole surface is covered, com- 
mencing at the first space each time. By this method the 
shading shows streaks of tint of different depths, but are almost 
unnoticeable if the divisions are taken quite small. 

The American method is most used, and is called shading hy 
softened tints. 

There are two ways of doing this : — 

1st. By mixing a small amount of dark wash at first, and 
starting as if you were to put on a flat tint, and then, by re- 
peated additions of clean water, going over a little more surface 
at each addition, gradually make the dark tint lighter until you 
are using almost pure water. 

2nd. Divide the surface into divisions, as in the French 
method, only not so many ; put on a flat medium tint on the 
first space, but, instead of taking up all the tint from the bot- 
tom edge of the surface, leave a slight amount, touch the brush 
to some clean water and apply it to the lower edge of the pud- 
dle, thus making a lighter tint, and bring down this new tint a 
short distance ; repeat this a few times until the tint has prac- 
tically no color. Be careful to remove the most of the tint from 
the brush each time before touching it to the clear water. This 
work must be done even quicker than the ordinary flat tint. 
Use as little tint or water in the brush as you can and not 
have it dry in streaks. Let this dry, and then repeat the proc- 
ess, commencing at the top and going over two spaces with the 
flat tint and softening off the lower edge, and so on, commenc- 
ing at the top each time. Usually the tint should be softened 
out in the length of one division. If this shading is done per- 
fectly, there will be a gradual change in the tint from beginning 
to end. 



CHAPTER IV, 



PROJECTIONS. 



25. Orthographic Projection, or Descriptive Geometry, 
is the art of representing a definite body in space upon two 
planes, at right angles with each other, by lines falling perpen- 
dicularly to the planes from all the points of the intersection of 
every two contiguous sides of the body, and from ail points of 
its contour. 

26. These planes are called coordinate planes, or the planes 
of projection, one of which is horizontal, and the other vertical. 
H and V, Fig. 1, represent two such planes and their line of 
intersection GL is called the ground line. 

27. We shall only take, in this book, just enough of the 
elementary principles of projections to enable the student to 
make working drawings of simple objects. 

28. Since solids are usually made up of planes, planes of 
lines, and lines of points, if we thoroughly understand the prin- 
ciples involved in the projections of points, we ought to be able 
to draw the projections of lines, of planes, and of solids. The 
only difference being that, with a large number of points, the 
student is liable to get them confused. To avoid this liability 
it is advisable, at first, to number the points of a solid and their 
corresponding projections as fast as found lightly in pencil, and 
erase them after the problem is finished. Do not try to draw 



66 PROJECTIONS. 

the object all at once, it is impossible ; one point at a time is 
all that can be drawn by anybody, and in this way the most 
complicated objects become simple, even though it may take a 
long time to complete the drawing. 

29. The projection of any point in space on a plane is the 
point at ivhich a perpendicular drawn from the given point to 
the plane pierces the plane. 

This perpendicular is called the projecting line of the point. 
Thus, in Fig. 1, a h is the projection of the point a on the plane 
H, and a t v of the same point on the plane V. These are called 
respectively the horizontal and vertical projections of the point 
a ; aa h is called the horizontal projecting line of the point a, 
and aa t v the vertical projecting line of the same point. 

The horizontal projecting line aa h is perpendicular, to H, by 
definition, the plane V is assumed perpendicular to H, hence 
aa h is parallel to the plane V, and aa t v is equal in length to 
a h b. Also the vertical projecting line, for the same reason, is 
parallel to H, consequently aa h is equal in length to a*b. 

From the definition it is readily seen that each point in a line 
perpendicular to a plane will have its projection on that plane 
in one and the same point ; hence one projection of a point 
does not definitely locate its position in space. 

30. From the preceding article the following principles may 
be noted : — 

First, the perpendicular distance from the horizontal projec- 
tion of a point to the ground line is equal to the perpendicular 
distance of the point in space from the vertical plane ; or, 
briefly, the horizontal projection of a point indicates the distance 
of the point in space in front of V, but it conveys no idea of its 
distance above H. 

Second, the perpendicular distance from the vertical projec- 
tion of a point to the ground line is equal to the perpendicular 



PROJECTIONS. 67 

distance of the point in space from the horizontal plane ; or, 
briefly, the vertical 'projection of a point indicates the height of 
the point in space above H, but it conveys no idea of its dis- 
tance in front of V. 

31. If from the points a t v and a h , Fig. 1, perpendiculars 
should be erected to each coordinate plane, they will intersect 
at the point a in space ; and as two straight lines can intersect 
at only one point, there is only one point in space which can 
have a h and af for its projections. Hence two projections of 
a point are always necessary to definitely locate its position in 
space. 

32. It is evident that it would be very awkward to make 
our drawings on planes at right angles to each other ; hence 
the vertical coordinate plane is supposed to be revolved back- 
ward about its line of intersection GrL with the horizontal 
plane until it forms one and the same surface with the hori- 
zontal plane, which may be considered to be the plane of the 
paper. 

In this revolution all points in the vertical plane keep the 
same distance from the ground line, and their relative positions 
remain unchanged. Thus, af revolves to a v , a v b being equal 
to afb, and as ajb was perpendicular to GL before revolution 
it will be so after revolution, and will form one and the same 
straight line with a h b. 

Therefore, the two projections of a point must always be on 
one and the same straight line, perpendicular to the ground line. 

Now, if we draw a line across our paper and call it GrL, all 
that portion in front of this line will represent the horizontal 
plane, and that portion behind it will represent the vertical 
plane, and the point a located as in Fig. 1 is represented by its 
projections on the plane of the paper as shown in Fig. 2. 



68 PROJECTIONS. 

33. A point situated upon either of the coordinate planes has 
for its projection on that plane the point itself and its other pro- 
jection is in the ground line. 

This is readily seen by referring to Fig. 1 ; c v and c h are the 
projections of a point on H, and rf* and d h of a point in V. 
These points are represented on the plane of the paper as shown 
by the same letters in Fig. 2. 

NOTATION. 

34. We will designate a point in space by a small letter, and 
its projections by the same letter with an h or v written above ; 
thus a h represents the horizontal and a v the vertical projection 
of the point a. This point may be spoken of as the point a, 
or as the point whose projections are a v a h . 

35. The horizontal coordinate plane will be designated by 
the capital letter H, the vertical by the capital letter V. 

36. Construction lines are those which are made use of sim- 
ply to obtain required results. They are not a necessary part 
of the drawing, and when left on a drawing are intended to 
show the individual steps taken. 

To this end the student is expected to ink in all construction 
lines illustrating the special subject in hand until especially 
directed not to do so. That is, while on the subject of projec- 
tions it is not desired to have the geometrical construction lines 
inked in, but only those which refer to projections. When on 
the subject of shadows only those which show how the shadow 
is found are required. 

These lines should be inked in with a light, short dash not 
more than T \ of an inch long, and as light as the student 
finds he can make easily. 

All lines representing the projections of single lines, or edges 
of planes or solids, if visible, are inked in with a full, continu- 



PROJECTIONS. 69 

ous line, a little heavier than the construction lines ; if invisi- 
ble, they should be made with short dashes the same length as 
the construction lines, but the same thickness as the visible lines, 
so as to distinguish them from the construction lines. 

The true length of a line when found should be inked in 
with a long and short dash, about the same thickness as the 
ordinary full line. When the true length of a line is given, it 
is put in like an ordinary construction line. 

Indicate an isolated point by drawing a small cross through it. 

37. In working drawings — which are practical applications 
of projections — horizontal projections are usually called plans, 
and vertical projections are called elevations. Therefore, they 
will be used synonymously throughout this book. 

38. The student should distinguish between the terms ver- 
tical and perpendicular. Vertical is an absolute term, and 
applies to a line or surface at right angles to the plane of the 
horizon while perpendicular is a relative term, and applies to 
any line or surface which is at right angles to any other line or 
surface. 

If one point is farther from V than another, the first is said 
to be in front of the second point. Hence, if I say that a line 
slopes downward, backward, and to the left, it signifies that the 
line occupies such a position that the lower end is nearer V 
than the upper, and also that it is on the left of the upper end. 
Fig. 6 shows the projections of such a line. 

PROJECTIONS OF STRAIGHT LINES. 

39. A straight line is determined by two points, therefore 
it is only necessary to draw the projections of each end of a 
straight line and join them, and we have the projections of the 
line. If the line is curved, it becomes necessary to draw the 
projections of several points and join them with a curve. 



70 PROJECTIONS. 

40. If we lay a fine wire, ab, Fig. 3, on a horizontal plane, 
and also parallel to a vertical plane, its horizontal projection 
will be the wire itself, that is, a h b h is its horizontal projection, 
equal in length to the wire itself, parallel to GL, and at a dis- 
tance from it equal to the distance of the wire from V. 

The vertical projection of the end a will be at a v in the 
ground line, and of b at b v also in the ground line, since each 
end is on H (Art. 9). Hence the vertical projection of the 
line will coincide with GL between a v and b v , and a v b v is its 
vertical projection. a v b v is equal in length to a h b\ hence is 
equal to the actual length of the wire. 

Now, suppose the wire to be revolved from right to left about 
a horizontal axis, through the end a, keeping the line parallel 
to V. If a pencil were attached to the end b at right angles 
to the line, so that its point touched V at b v , it would trace a 
circular arc b v c v d v , etc. (of which a v is the centre and a v b v , or 
the true length of the line, is the radius) on the vertical plane 
as the wire is revolved ; the end a would, of course, not move. 
After the wire has been revolved through an angle of 30° its 
vertical projection will be at a v c v , and it must be equal in length 
to the real length of the wire. a h will be the horizontal pro- 
jection of the fixed end of the wire. Since the wire is parallel 
to V, every point in it is at the same distance from V, hence 
their horizontal projections must all be the same distance from 
GL, that is, in a line parallel to GL ; the horizontal projec- 
tion of the end c must also be in a line through c v perpendicu- 
lar to GL, heDce at c h where this parallel and perpendicular 
intersect ; a h c h , then, is the horizontal projection of the wire 
after it has been revolved through an angle of 30°. 

For the same reason a v d v and a h d h are the two projections 
of the wire after being revolved through an angle of 45°. 

Similarly, a v e v and a h e h are its projections after revolving 
through an angle of 60°. 



PROJECTIONS. 71 

When the line has been revolved through 90° it becomes per- 
pendicular to H, and its vertical projection is a v f v , perpendicu- 
lar to GL, and its horizontal projection is a point, a h , as might 
have been seen from the definition of the projection of a point. 

41. The same reasoning applies if we take the wire lying 
against V and parallel to H and revolve it about a vertical axis 
through the end a, as in Fig. 4. The projections are given for 
the line lying against V, and making angles of 30°, 45°, 60°, 
and 90° with V, being parallel to H in each position. 

42. The following principles may be noted from the pre- 
ceding articles : — 

1st. A line situated in either plane is its own projection on 
that plane, and its other projection is in the ground line. 

2nd. If a right line is perpendiadar to either plane of pro- 
jection, its projection on that plane will be a point, and its pro- 
jection on the other plane will be perpendicular to the ground 
line and equal in length to the given line. 

3rd. When a line is parallel to either coordinate plane its 
projection on that plane will be parallel to the line itself, and 
equal to the actual length of the line in space, and its projection 
on the other plane will be parallel to the ground line. 

4th. If a line is parallel to both planes, or to the ground line, 
both projections will be parallel to the ground line. 

bth. If a line is oblique to either coordinate plane its pro- 
jection on that plane will be shorter than the actual length of the 
line itself. 

6th. If a line is parallel to one coordinate plane and oblique 
to the other, its projection on the plane to which it is parallel is 
equal to the true length of the line in space, and the angle which 
this projection makes with the ground line is equal to the true 
size of the angle the line in space makes with the plane to which 
it is oblique. 



72 PROJECTIONS. 

1th. The projection of a line on a plane can never be longer 
than the line itself. 

&th. If a point be on a line its projections will be on the 
projections of the line. 

9th. If a line in different positions makes a constant angle 
with a plane its projections on that plane will all be of the same 
length, without regard to the position it may occupy relative to 
the other plane. 

43. If two lines intersect in space their projections must also 
intersect, and the straight line joining the points in which the 
projections intersect must be perpendicular to the ground line ; 
for the intersection of two lines must be a point common to 
both lines, whose projections must be on the horizontal and ver- 
tical projections of each of the lines, hence at their intersec- 
tions respectively. 

44. If two lines are parallel in space their projections upon 
the vertical and horizontal planes will be parallel respectively. 
If one projection only of two lines are parallel, the lines in 
space are not parallel. 

45. Any two lines drawn at pleasure, except parallel to each 
other and perpendicular to the ground line, will represent the 
projections of a line in space. 

46. Prob. 1. To draw the projections of a line of a defi- 
nite length and occupying a fixed position in space. 

Let it be required to draw the projections of a line 1" long, 
which makes an angle of 30° with H, and whose horizontal 
projection makes an angle of 45° with GL, the lower end of 
the line being -J" above H and J" in front of V. 

It is first necessary to place the line in such a position that 
its true length and the true size of the angle it makes with one 
of the coordinate planes are shown, and these are only shown 



PROJECTIONS. 73 

when it is parallel to one of the coordinate planes. In this case 
it must be placed parallel to V. a v and a h , Fig. 5, are the pro- 
jections of one end of the line, a v being J" above GL and a h 
J»" below it. Through a v draw a v bf at an angle of .30° with 
GL and 1" long ; through a h draw a h b t h parallel to GL, b t h be- 
ing found by dropping a perpendicular from b v (Art. 40).. The 
two projections of the line, when parallel to V, are thus found 
to be a v b? and a h b\ 

Now let us suppose the end a of the line to be fixed and the 
whole line to be revolved through an angle of 45° about a ver- 
tical axis through this point, the line keeping the same angle 
with H. The horizontal projection will not change in length 
(Art. 42, 9th), but will move through an angle of 45°, and will 
be found at a h b h . It is evident that in this revolution, so long 
as the angle with H does not change, every point in the line will 
remain at the same height above H. The point a does not move, 
being in the axis. We have seen that b t h moves to b h ; b v must, 
therefore, be somewhere on a perpendicular through b h , and, 
since the points do not change their heights, it must also be on 
a line through b t v parallel to GL, hence at their intersection b v . 
Join a° and b v and we have a v b v and a h b h as the required pro- 
jections of the line. 

47. If this line were revolved through 15° more, the point 
U 1 would go to c h , and b" to c v , and a v c v and a h c h would be the 
projections of the line making an angle of 30° with H, and 
whose horizontal projection made an angle of 60° with GL. 

If it were revolved still 30° more, the two projections would 
be a v d v and a h d\ each being perpendicular to the ground line. 
When a line is in this position, i. e., has its two projections in 
a line perpendicular to GL, it is said to be in a profile plane, 
a profile plane being understood to be one that is perpendicular 
to both V and H. 



74 PROJECTIONS. 

When a line is oblique to only one of the coordinate planes 
it is said to make a simple angle ; when it is oblique to both of 
them it is said to make a compound angle. 

48. If the angle that the line made with V had been given, 
it would have been necessary to have first placed the line par- 
allel to H, and then to have revolved it about an axis through 
one end perpendicular to V, in which case the length of the 
vertical projection would not change, and the points would not 
change their distances from V. 

In Fig. 6, a v b v and a h b h are the two projections of a line 1" 
long, making an angle of 45° with V, and whose vertical pro- 
jection makes an angle of 60° with GL. The principles and 
explanation for this construction are the same as for Prob. 1, 
if the horizontal and vertical planes are supposed to be inter- 
changed. 

49. Prob. 2. To find the true length of a line given by its 
projections, and the angle it makes with either plane of projec- 
tion. 

Let a v b v and a h b h , Fig. 7, be the projections of the given line. 
The true length is only shown when it is parallel to one of the 
coordinate planes, hence this line must be revolved about an 
axis through either end until it is parallel to one of the planes. 
If it is revolved about a vertical axis through a until it is par- 
allel to V, the point a does not move, b h moves to bj 1 , b v is found 
at bj (where a perpendicular through b t h intersects a horizontal 
through b v ), and cPhf is the true length. Also, the angle which 
a v b* makes with GL is equal to the true size of the angle the 
line makes with H. 

If it had been required to find the angle this line made with 
V, it would have been necessary to have revolved the line about 
a horizontal axis until it became parallel to H. Assuming the 



PROJECTIONS. 75 

axis through the end b, Fig. 7, a v moves to a*, a h to a ^, and 
a*b h is the true length of the line (which of course should equal 
a v b"), and the angle it makes with GL is equal to the true size 
of the angle the line makes with V. 

50. Note. The angles which the vertical and horizontal 
projections of a line make with GL are greater than the angles 
which the line in space makes with H and V respectively, except 
when the line is parallel to one of the planes. 

PROJECTIONS OF SURFACES. 

51. Plane surfaces are bounded by lines, therefore the prin- 
ciples which govern the projections of lines are equally appli- 
cable to these surfaces. 

52. If we suppose a rectangular card abed, Fig. 8, placed 
with its surface parallel to V and perpendicular to H, each edge 
being parallel to V, it will be projected on V in a line equal and 
parallel to itself, hence the true size of the card itself is shown 
in vertical projection. Two of the edges, ab and cd, being per- 
pendicular to H, are projected on that plane in the points a h 
and d h respectively. The other two edges, ad and be, are par- 
allel to H as well as V, hence they will be projected in their 
true length on H. and, since one is vertically over the other, 
they will both be horizontally projected in the same line a h d h . 

Now, if the card is revolved about one of its vertical edges 
as an axis, like a door on its hinges, the vertical edge which 
coincides with the axis does not move ; the other vertical edge 
moves in the arc of a circle. The horizontal projection of the 
card will still be a straight line of the same length as before. 
Let the card be revolved through 60° ; a h does not move ; d h 
moves in the arc of a circle, of which a h is the centre and a h d h 
the radius, to df 1 -, a h d h is the horizontal projection of the card 
in its new position ; the vertical projection of the edge cd in 



76 PROJECTIONS. 

this position is found at c"^", vertically above dj 1 , and a v b v c v d t v 
is the vertical projection of the card after being revolved 
through an angle of 60°. 

If the card should be revolved through 30° more, i. e., 90° 
in all, its surface will be at right angles with both coordinate 
planes, and its two projections will be found at a°b v and a h e h , in 
one and the same straight line perpendicular to GL. 

53. If the card be placed on H, with one of its edges par- 
allel to V, a h b h c h d h , Fig. 9, will be its horizontal and a v b v its 
vertical projection. If this card be revolved about one of its 
edges which are perpendicular to V as an axis, like a trap-door 
on its hinges, through an angle of 30°, a v b® and a h b^c^d h will 
be its two projections. If it be revolved through 60° more, or 
90° in all, its projections will be a v e v and a h d h , which are just 
the same as a v b v and a h e h in Fig. 8, as they should be, since the 
cards are the same size in the two figures and they occupy the 
same relative position in each, i. e., they are in a profile plane. 

54. The following principles may be noted : — 

1st. When a plane surface is perpendicular to another plane 
its projection on that plane will be a line. 

2nd. When a plane surface is parallel to either coordinate 
plane, its projection on that plane will be equal to the true size 
of the surface and its other projection will be a line parallel to 
GL. 

3rd. When a plane surface is perpendicular to one plane and 
oblique to the other, the angle which its projection on the plane to 
which it is perpendicular makes with the ground line is equal to 
the angle the surface in space makes with the plane to which it is 



4dh. If a plane surface, in different positions, makes a con- 
stant angle with a plane, its projections on that plane will all be 
of the same size. 



PROJECTIONS. 77 

55. Prob. 3. To draw the tvjo projections of a plane sur- 
face, or card, of a certain size, and making a compound angle 
with the coordinate planes. 

Let the card be of the size shown in Fig. 9, and suppose it 
to make an angle of 30° with H, and its horizontal projection 
an angle of 45° with GL. 

Draw the horizontal projection aJW'd* of the card equal to 
its true size ; a v b v will be its vertical projection. Revolve a v b v 
through an angle of 30° to a v b t v , and a v b" will be the vertical 
projection of the card when it makes an angle of 30° with H 
and is perpendicular to V ; a h b h c t h d h is its corresponding hori- 
zontal projection. 

The angle with H is still to be 30° after the card has been 
revolved to its desired position, hence its horizontal projection 
will be the same size. Therefore, make aPbJ'cW 1 , Fig. 10, 
equal in size to a^bf-cH 71 , Fig. 9, and making the desired angle 
with GL. hi this revolution, as long as one edge rests on H 
and the angle remains constant with H, every point keeps the 
same height above H, therefore the vertical projections of a h 
and d h , Fig. 10, must be found at a v and d° ; also of b h and c/ 4 
at b t v and c", whose heights above GL are equal to the height 
of b% Fig. 9, above GL. 

The other parallelograms in Fig. 10 represent the projections 
of the same card at different angles with H, the horizontal pro- 
jections making the same angles with GL. 

56. a h b h c h d h and a v d v b v c v , Fig. 11, represent the projections 
of the same card when it is lying on H with one of its diago- 
nals parallel to V, and a v d t v b t v c t v and aPb^cHf 1 are its projec- 
tions after being revolved about an axis perpendicular to V 
through the corner a through an angle of 45°. Fig. 12 repre- 
sents the projections of this card when, besides making an angle 
of 45° with H, the horizontal projection of the diagonal makes 



78 PROJECTIONS. 

an angle of 30° with GL. The steps to obtain this are exactly 
the same as in Figs. 9 and 10, hence the explanation will not 
be repeated. 

57. Prob. 4. To draw the projections of a regular pentag- 
onal card, the diameter of the circumscribed circle being given, 
in two positions. 1st, when it is perpendicular to V and making 
an angle of 60° with H, one of its edges being perpendicular to 
V ; 2nd, when, besides making an angle of 60° with H as in 
1st position, it has been revolved through an angle of 45°. 

The pentagon must first be drawn in its true size and posi- 
tion ; a^b^c^dH 1 ^, Fig. 13, equal to the actual size of the card, 
is its horizontal projection, c t h d^ being perpendicular to GL, 
and a v b^cj is its vertical projection. Revolve a v b^c^ through 
an angle of 60° to a v b v c v ; each point moves in the arc of a cir- 
cle with a as a centre, and a v b v c v will be the vertical projection 
of the card when in the 1st position asked for ; a h b h c h d h e h is 
its corresponding horizontal projection. 

For the 2nd position revolve the plan just found through 45° 
to the position a h b h , etc., shown in Fig. 14 ; a v b v c v d v e v will be 
the corresponding vertical projection. 

* 58. Cards of any shape and size, and occupying any position, 
may be drawn in the same way, care being taken to locate one 
point at a time. 

59. If the angle the card made with V had been given, it 
would have been necessary to have first placed the card par- 
allel to V and then to have revolved it, through the angle it 
made with V, about an axis perpendicular to H, in which case 
the length of the horizontal projection, which is a straight line, 
would not change, and the several points would not change 
their respective distances from H. 



PBOJECTIONS. 79 

In the second revolution, which changes the angle with the 
coordinate planes from simple to compound, the vertical projec- 
tion must be revolved and the corresponding plan found. The 
angle with V being the same the vertical projection does not 
change its size ; the distances of the points in front of V remain 
the same after revolution as before, hence are found at the same 
distances from GL respectively. 

60. Prob. 5. To draw the projections of a circular card 
making a compound angle with the coordinate planes. 

Let the diameter of the card be given, the aDgle it makes 
with V, and the angle through which the vertical projection is 
to be revolved. 

A circle may be considered as a polygon of an infinite num- 
ber of sides, hence we can take as many points as we please on 
the circumference of the circle, and each one moves according 
to the principles just described. 

Place the card parallel to V ; a circle, a r b*c*\ etc., Fig. 15, 
equal to the actual size of the given circle, is its vertical pro- 
jection, and aPb^cJ 1 , etc. is its horizontal projection. 

Revolve the card through the required angle about a vertical 
axis through a ; a h b h c h , etc. is its horizontal, and a, v b v c v , etc. is 
its vertical projection. 

The card is to be revolved through a certain an^le, still 
keeping the same angle with V. The size of the vertical pro- 
jection will, therefore, not change. Hence, revolve the vertical 
projection found in Fig. 15 through the required angle to the 
position a v b v c v , etc., Fig. 16. None of the points change their 
distance from V, consequently a h b h c h , etc. is the horizontal pro- 
jection of the card, found as in the last problem. 

61. Fig. 17 shows a somewhat shorter method of drawing 
the projections of a circular card making a simple angle with 



80 PROJECTIONS. 

the coordinate planes; in this case it makes an angle of 30° 
with V and is perpendicular to H. a h b h c h , etc., making 30° 
with GL, is its horizontal projection. Suppose the card re- 
volved about its horizontal diameter ae until it is parallel to H. 
It will then be shown in its true size at a h bc, etc. ; bl/ 1 , cc h , 
etc. will show the actual distances of the points b, c, etc. from 
the horizontal diameter ; a v e° will represent the vertical projec- 
tion of the horizontal diameter about which the card is revolved. 
Of course, b v must be found somewhere in a line through b h , 
perpendicular to GL, therefore, lay off tb° equal to b b* 1 , and b v 
is a point of the required vertical projection ; c v s is made equal 
to c c h , d v r to dd h , etc. Other points may be found in the same 
way. 

It is evident that nb h , mc h , etc. are respectively equal to 
hb h , cc h , etc, hence it is only necessary to revolve the semi- 
circle and the distance bb h is laid off on both sides of the diam- 
eter a v e v , giving the two points If and n v . 

PROJECTIONS OF SOLIDS. 

62. A cube is a solid bounded by six equal faces, and when 
it is placed so that two of its faces are parallel to H, and two 
others parallel to V, its two projections are a v b v e v f v and a h b h c h d h , 
Fig. 18. The top and bottom, being parallel to H, are hori- 
zontally projected in one and the same square, a h lfic h d h , which 
is, of course, equal to the exact size of these faces, and their 
vertical projections are a v b v and e v f v respectively (Art. 30-2nd) ; 
the front and back faces being parallel to V are vertically pro- 
jected in one and the same square, a v b°e v f v , which is also equal 
to the exact size of these faces, and their horizontal projections 
are a h b h and c h d h respectively (Art. 30-2nd) ; the left and right 
hand faces are perpendicular to both V and H, therefore their 
vertical projections are a v e v and b v f v , and their horizontal pro- 
jections are a h d h and b h c h respectively (Art. 30-1 st). 



PROJECTIONS. 81 

The plan shows two dimensions of the cube, the length and 
breadth, and the elevation two dimensions, the length and thick- 
ness ; therefore, the three dimensions of the solid being shown 
in their true size in the two projections the object is com- 
pletely represented. In this case it does not matter which pro- 
jection is drawn first, as they each show two dimensions in their 
true size. 

63. Shade Lines. In outline drawings it is customary to 
put in shade lines, i. e., lines heavier than the others ; they give 
relief to the drawing, and, ivhen properly placed, are of assist- 
ance in reading it. 

Shade lines, or edges, are those edges which separate light 
from dark surfaces. 

The rays of light are generally assumed to come from over 
the left shoulder in the direction of the diagonal of a cube, the 
person supposed to be facing the cube, and the cube to be in 
the position shown in Fig. 18. That is, the ray of light enters 
the cube at the upper, front, left-hand corner, whose projections 
are a v and a h , and leaves it at the lower, back, right-hand corner, 
whose projections are f° and c h ; the diagonal joining these 
points will represent the actual direction of the conventional 
ray of light, and its projections a v f v and a h c h are the projec- 
tions of this ray. The different rays of light are all supposed 
to be parallel to each other. 

It is evident from the figure that both projections of the rays 
of light make angles of 45° with GL. The student should 
distinctly understand that, although the projections of the ray 
of light make 45° with GL, the actual angle it makes with V 
or H is quite different. To find this angle apply the princi- 
ples of Art. 25 to Fig. 18, and we get a = 35° 15' 52" as its 
actual size. 

G4. In the cube, Fig. 18, the top, front, and left-hand faces 



82 PROJECTIONS. 

are light, and the bottom, back, and right-hand faces are dark, 
and the shade lines are, therefore, e v f v , which separates the 
front from the bottom, b v f°, which separates the front from the 
right-hand face, b h c h , which separates the top from the right- 
hand face, and c h d h , which separates the top from the back face. 
The two other shade edges which separate the left-hand face 
from the back and bottom faces are not seen in either projec- 
tion, since the top and front edges of the left-hand face are in 
the same plane and nearer the eye. 

It is for this same reason that the shade lines mentioned 
above are seen only in one projection, i. e., e v f v is a shade line ; 
it is seen in elevation, but in plan is hidden by the upper front 
edge of the cube a v b v -a ], b h . 

65. It will be noticed that the right-hand and lower edges 
are shaded in elevation, and the right hand and upper in plan. 
From this many draftsmen have adopted the arbitrary rule to 
shade the right-hand and lower lines in elevation, and the right- 
hand and upper in plan. This rule is really applicable in but 
few cases, except when the object is of rectangular section and 
so placed that its surfaces are perpendicular to one or both of 
the coordinate planes. 

Other draftsmen shade the right-hand and lower lines in 
both plan and elevation. This is also applicable only in the 
cases stated above, besides being obliged to change the direction 
of the ray of light in plan and elevation, or imagine the object 
revolved while the ray of light remains fixed. 

Others still follow the rule given in Art. 63, except that they 
only put in shade lines where the dark portion of space adja- 
cent to the line in question is visible, and they change the direc- 
tion of the ray of light, or, what is practically the same thing, 
revolve the object. By this method the plan and elevation are 
shaded alike, but the right-hand line is not necessarily a shade 
line, while the left-hand line is necessarily not a shade line. 



PROJECTIONS. 83 

The method taken up in this book, and the one it is expected 
that the student will follow in this course, is not given because 
it is the one most generally in use, or because it is the easiest, — 
quite the contrary ; but because it is, in the opinion of the 
writer, the only method which can be followed consistently 
throughout a course of projections, shadows, isometric, and 
working drawings. 

66. Prob. 6. To draw the two projections of a right square 
prism with its base on H and its vertical faces oblique to V. 
Fig. 19. 

It is evident that the vertical faces will not be projected on 
V in their true size, the height only of the prism being shown 
in its real length, and also that the two ends, being parallel to 
H, will be projected on that plane in their true size. Hence, 
draw the square a h b h c h d h equal to the ends of the prism, with 
its edges making the same angles with GL as the vertical faces 
make in space with V. Through the corners a h , b h , c h , and d h , 
draw the perpendiculars a v e v , b v f v , etc., making each equal in 
length to the height of the prism. a v b v c r d v will be the vertical 
projection of the upper base. e v f v m v n v of the lower base, and 
a v c v m v e v the vertical projection of the whole prism. 

In this case the top and two front faces are light, while the 
bottom and two back faces are dark ; hence, in plan the lines 
a h d h and c h d h , which separate the top from the two back faces, 
are shade lines ; in elevation they are behind the two front edges 
ab and be, consequently are not seen. The element ae separates 
the left front from the left back face, hence is a shade line, and 
a v e V ) its vertical projection, is accordingly made heavy ; its hori- 
zontal projection is simply a point ; the element cm also sepa- 
rates a light from a dark surface, and its visible projection c v m v 
should be made heavy. The edges ef and fm separate the two 



84 PROJECTIONS. 

front faces from the bottom, and their visible projections e v f v 
and f v m v are also made heavy. 

r 

67. Prob. 7. To draw the two projections of a right regu- 
lar pentagonal prism standing with its base on H, with none of 
its faces parallel to V. Fig. 20. 

Here, as in the last problem, it is necessary to draw the hori- 
zontal projection first, as it shows the pentagonal end in its true 
size and position. a h b h c h d h e h is its horizontal projection, and 
a v d v n v f v , found as in the last problem, is its vertical projection. 

From the shade lines shown in the figure it will be noticed 
that the line cd, which separates the light top from the dark 
right-hand face, is visible in both projections, hence it is made 
heavy in both. 

A prism of any number of sides standing on H can be drawn 
in the same manner. 

68. Prob. 8. To draw the two projections of a right regu- 
lar hexagonal prism with its axis perpendicular to V. Fig. 21. 

Here it is necessary to draw the vertical projection first, and 
construct the horizontal projection from it according to the 
principles noted in the last two problems. 

69. From an inspection of Figures 18 to 21 it is evident that 
the 45° triangle can be used to determine positively the light and 
dark faces only when these faces are perpendicular \ or nearly 
so, to one or both of the coordinate planes. 

In Fig. 18 the triangle can be used in both plan and eleva- 
tion, since every face is perpendicular to at least one of the 
coordinate planes. In Figs. 19 and 20 the faces are perpen- 
dicular to H only (except the top, which is, of course, known to 
be light), hence the 45° triangle can only be u^ed in the plan. 



Fig. 5 Fig. 6 




PROJECTIONS. 85 

In Fig. 21 the faces are perpendicular to V only (except the 
front end, which must be light), hence the triangle can only be 
used in elevation. 

70. Prob. 9. To draw the two projections of a right cylin- 
der standing on its base. Fig. 22. 

Its horizontal projection will be a circle equal to the end of 
the cylinder (Art. 42). Its vertical projection will be repre- 
sented by a v b v d v c v , a v c v being the vertical projection of the top, 
b v d v of the bottom, a v b v and c v d v of the extreme left and right 
hand elements, or the contour lines as they are called. 

By applying the 45° triangle it is evident that the shade line 
in plan will be the half circle mW* between the points where 
the triangle is tangent to the circle. The element cd is not a 
shade line, as it does not separate a light from a dark surface. 
The shade element would be or, but as it is not drawn of course 
it cannot be shaded. 

The bottom of the cylinder is dark, and strictly the line b r r" 
would be heavy, leaving the portion r v d° as light ; but, since it 
is practically impossible to stop the shade line at the point r° 
and make a good-looking line, I should disregard this short piece 
and shade the line the whole length. Similarly, on top there 
will be a short piece between o v and c v that would strictly be 
shaded, but for the same reason I would disregard this and 
make the whole top a light line. 

The absence of the shade line c v d v in the vertical projection 
enables us to tell at once that the object is cylindrical even 
before we look at its plan. 

A cone is like a cylinder, except that its elements all intersect 
in a common point called the vertex, while in the cylinder they 
are all parallel. Fig. 51 represents the two projections of a cone. 



86 PROJECTIONS. 

71. Prob. 10. To draw the two 'projections of a right regu- 
lar hexagonal pyramid with its base resting on H. Fig. 23. 

As we look directly down upon a pyramid, in this position 
we shall see all of its sloping faces, and consequently the edges 
which separate these faces. We shall also see the edges which 
separate these sloping faces from the base, i. e., the outline 
of the base. In this case the base is parallel to H, hence its 
outline is shown in plan equal to the real size of the base. 
Therefore, the regular hexagon, a h b h c h d h e }i f h , is the horizontal 
projection of the outline of the base. Now, since in a right 
pyramid the base is perpendicular to the axis, it will be easily 
seen that the horizontal projection of the vertex of the pyramid 
must be at o h , the centre of the hexagon. 

Drawing lines from o h to each corner of the hexagon, the 
horizontal projection of the pyramid is completed. The points 
in the base are, of course, vertically projected in GL, the ver- 
tex at d° at a distance above GL equal to the altitude of the 
pyramid. Joining 6° with a v ,h v ,c v , etc. we have its vertical 
projection. 

The shade lines of a pyramid are not found directly by means 
of the 45° triangle, as we have been able to do previous to this, 
on account of the faces not being perpendicular to either coor- 
dinate plane. If we try to use the triangle as in the case of 
the prism, we would have said that the three faces, f h o h a h , 
a h o h b h , and b h o h c h were light, and the three remaining faces 
dark, but this is not the case. For let us suppose that the alti- 
tude of this pyramid is so small that each of the faces of the 
pyramid makes an angle with H less than 35° 16' (the angle 
the ray of light makes with H). It is evident that all of the 
sloping faces will be light, and the bottom being dark the shade 
lines would go entirely round the base. Now, if we consider 
the altitude to increase, we shall soon reach the point when the 



PROJECTIONS. 87 

face o h d h e h will become dark, all of the rest remaining light, 
and the shade line would change from d h e h to e h o h and o h d h . If 
the altitude be still further increased, we next get the case,shown 
in the figure where the la,cef h o h e h becomes dark, and the shade 
lines would change from e h f h and e h o h to f h o h . If the altitude 
should be still further increased, the face c h o h d h would presently- 
become dark also. 

Of course the other three faces would never become dark 
while the pyramid rested on its base, even if the vertex were 
extended to infinity, in which case we should simply have a 
prism. In cases like this, or where any surface is oblique to 
both V and H, it is necessary to fiud the shadow of the object, 
thus determining which surfaces are light and which are dark. 

72. In Fig. 24 a v b v f v e v is the elevation and a h b h c h d h is the 
plan of a rectangular prism, with two of its faces parallel to each 
of the coordinate planes. The plan shows its length and width, 
and the elevation its length and thickness. If a side elevation 
is desired, it will show the width and thickness. To get this 
the object must be projected onto a plane at right angles to the 
two coordinate planes, i. e., the profile plane, and this plane 
revolved about its intersection with V, as an axis, to coincide 
with V. POR is such a plane, resting against the end of the 
prism, PO being its intersection with V and OR its intersection 
with H. 

In this revolution none of the points change their heights 
above H, nor their distances from the axis PO, hence the rect- 
angle hfcffjm* will represent this side elevation, it being of 
course the same height above GL that the front elevation is, 
and the distance that c *m* is from the axis PO will be equal to 
the distance the back of the prism is in front of V. 

The shade lines in the end elevation are shaded the same way 



88 PROJECTIONS. 

as in the front elevation ; the ray of light is supposed to come 
from over the person's left shoulder when he is facing tne pro- 
file plane, i. e., the vertical projection of the ray of light is the 
same for all elevations. 

73. Prob. 11. To draw the plan and two elevations of a 
square prism with its axis parallel to and at a definite distance 
from both V and H, all of its faces being oblique to both V and 
H. Fig. 25. . 

The end elevation is the only view of the prism which shows 
one of its surfaces in its true size and position relative to the 
coordinate planes, hence this view must be drawn first. 

Locate the point d° at a perpendicular distance above GL 
equal to the height of the axis above H ; through o v draw two 
lines at right angles to each other, making angles with GL equal 
to those made by the long faces of the prism with H respect- 
ively ; lay off on each of these lines, on both sides of o v , a dis- 
tance equal to half the side of the square ; through these points 
draw lines parallel to the lines through o v , and the square afbjc* 
d t v thus formed will be the end elevation of the prism in its cor- 
rect position. 

To locate the axis the correct distance from V draw POR, 
which represents the profile plane, perpendicular to GL and at 
a horizontal distance to the right of o v , equal to the distance of 
the axis in front of V. 

In the last problem the end view was constructed from the 
plan and front elevation ; in this problem we construct the 
plan and front elevation from the end view by simply reversing 
the steps. 

The horizontal lines, a v e v , b v f v , c v m v , and d v n v , drawn through 
a*, ft®, c®, and d v respectively, and each equal in length to the 
length of the prism, will be the front elevation of the different 



PEOJECTIONS. 89 

elements of the prism ; joining these ends the front elevation 
of the prism is complete. 

From O along OR lay off Om h , Of h , etc. equal to the hori- 
zontal distances of c r , & w , etc. from PO. Through these points 
m h ,f h , etc. draw the horizontal lines m h c h ,f h b h , etc., each equal 
in length to the length of the prism, and joining the ends the 
plan of the prism, a h c h m h e h , will be complete. 

The long faces of the prism being perpendicular to V in the 
end view, the shade lines for that view may be found directly 
by using the 45° triangle, as shown by the arrows. In revolv- 
ing the prism from the position shown in end view to that in 
the front view, the front and back ends change from light to 
dark and from dark to light respectively, but the long faces are 
light or dark in the front view and plan according as they are 
light or dark in the end view (provided the projection of the 
right-hand end is represented, which will be seen to the left of 
the front view). 

74. Prob. 12. To draw the two projections of a regular 
pentagonal prism, with its axis parallel to H and oblique to V, 
and its lower left-hand long face making a definite angle with H. 
Fig. 26. 

Here, as in the last problem, it is necessary to draw the view 
of the end of the prism when its axis is perpendicular to V, so 
as to show it in its true size and position. afb*c*d*e» is its 
end view, the edge a*bf making an angle with GL equal to the 
angle the lower left face makes with H. In the last problem 
the prism was revolved through an angle of 90° to its actual 
position, but in this it is revolved through a smaller angle. 
The steps being otherwise just the same, the explanation will 
not be repeated. 

The shade lines in this case may also be found, as in the last 
problem, by using the 45° triangle on the end view. 



90 PROJECTIONS. 

75. Let us dow suppose a case where the edge <?*«?*, Fig. 
26, makes an angle of 40° with H and in the same direction. 
It is evident that when the axis of the prism is perpendicular 
to V the surface which is projected in c*d* will be light. Now, 
if the prism be revolved through 45° so that its axis makes an 
angle of 45° with V, in the same direction as shown in Fig. 26, 
it is also evident that the surface, which makes an angle of 40° 
with H, will now be dark, and the shade lines would therefore 
change. If the prism be revolved through 45° more in the 
same direction, its axis would be parallel to V, and the surface 
in question would then be light. That is, the surface when 
perpendicular to V would be light, but as it was revolved par- 
allel to H, at some intermediate position before it had revolved 
45°, it would become dark, changing to light again at some 
intermediate position between 45° and 90° of revolution. 

The same thing would occur if the face in question made any 
angle with H between 35° 16' and 45°. The foregoing reason- 
ing would apply equally well to the under face ajh*, except that 
this one would be dark where the corresponding upper one 
would be light. 

Therefore, if a surface of a prism, as in the last problem, 
makes an angle with H between 35° 16' and 45°, that surface 
becomes doubtful in all its positions when the axis of the prism is 
oblique to V, and the shadow of this surface would have to be 
cast to determine positively whether it is light or dark. 

If the surfaces make angles with H, not included between 
the above limits, the 45° triangle on the end view would deter- 
mine the light and dark surfaces for all the oblique positions 
of the prism, as well as when the axis is perpendicular to V or 
parallel to V and H. 

76. Fig. 27 represents the two elevations and plan of a hol- 
low cylinder whose axis is parallel to V and H. Here the end 



PROJECTIONS. 91 

elevation would naturally be drawn first, as in the last two prob- 
lems, but it is not strictly necessary, as both of its projections, 
when parallel to V and H, are the same, and the distance apart 
of the contour elements is equal to the diameter of the base. 

The student should note carefully the shade lines in the figure, 
especially in the end view. 

77. Fig. 28 shows the plan and two elevations of a pile of 
blocks. The lower one is a rectangular prism, the second one, 
which rests on the first, is the frustum of a square pyramid, and 
the top one is a square pyramid. In this case it is necessary 
to draw the plan first and construct the two elevations from it, 
according to the principles already explained. 

It will be observed that the group is considered as solid in 
putting in shade lines, i. e., the edges which represent the peri- 
meter of the base of the pyramid, for example, are considered 
as separating the sloping faces of the pyramid from the top 
surface of the frustum on which it rests, and not from its base, 
as in Fig. 23. Compare the shade lines of the pyramids in 
Figs. 23 and 28. 

Since it is customary to tint drawings in which shadows are 
cast, shade lines would not be put in on the same drawing. 
Therefore, it is advisable to disregard the shadows altogether in 
putting in shade lines on line drawings. 

78. Prob. 13. To construct the projections of a right hex- 
agonal prism when its axis makes a compound angle with the 
coordinate plane, the angle it makes with H, the angle the hori- 
zontal projection makes with GL, and the size of the prism being 
given. Figs. 29, 30, and 31. 

There are three distinct steps necessary in the construction 
of this problem. 

Since the axis is oblique to both planes the prism cannot be 



92 PROJECTIONS. 

drawn in the required position directly, but must be placed in 
such a position as will show two dimensions. Here the first 
step is to draw the two projections of the prism when its axis 
is perpendicular to H and its faces make the required angles 
with V (Art. 67). Fig. 29 represents the projections of the 
prism in this position. 

Next draw the projections of the prism after it has been 
revolved in the proper direction, so that the axis makes the 
correct angle with H, but is still parallel to V. Fig. 30 repre- 
sents the projections of the prism in this position. Since the 
prism keeps a constant angle with V, its vertical projection 
does not change its size (Arts. 42-9th and 54-4th). Hence, 
make the vertical projection in Fig. 30 the same size as in 
Fig. 29, with its position changed so that the elements make 
the correct angle with GL. In this revolution no point changes 
its distance from V. Therefore, to construct the horizontal 
projection of the prism in this second position, draw through 
each corner, as a v , af, etc., Fig. 30, lines perpendicular to GL 
until they intersect horizontal lines drawn through the corre- 
sponding points, as a h in Fig. 29. This completes the second 
step. 

In the final position the axis is to make the same angle with 
H as it does in the position just drawn. Hence the horizontal 
projection must be the same size, however much it may be 
revolved. In this revolution no point changes its height above 
H. Therefore, draw the plan of the prism, Fig. 31, the same 
size as that in Fig. 30, only changing the angle the elements 
make with GL the required amount. Then from each corner 
of this plan, as a h ,a t h , etc., draw perpendiculars to GL until 
they intersect horizontal lines drawn through the corresponding 
points, a v ,a,v, etc., in Fig. 30. Joining these corners the ver- 
tical projection is completed. 






PROJECTIONS. 93 

If the angle the prism made with V and the angle the ver- 
tical projection made with GL had been given, the principles 
would have been just the same, only you would have first drawn 
it with its axis perpendicular to V, then revolved it about 
a vertical axis until it made the required angle with V; in 
this case the plan does not change its size ; and, lastly, revolved 
the vertical projection last found through the proper angle and 
constructed the corresponding plan. 

The shade lines in Figs. 30 and 31 can only be determined 
positively by casting the shadows of the doubtful surfaces. 
Frequently it is possible to tell which are the light and which 
the dark surfaces without casting the shadows by conceiving 
the solid in its position in space together with the ray of light, 
but for the average student it would be little better than a guess 
until he has had considerable practice in finding shadows. 

Before taking up these cases where it is necessary to cast the 
shadow in order to determine the shade line, it will be neces- 
sary to take up so much of the subject of shadows as will ena- 
ble the student to find the shadow of an ordinary object on the 
two coordinate planes. 

79. Prob. 14. To construct the projections of a right hep- 
tag onal pyramid when its axis makes a compound angle with the 
coordinate planes, the angle it makes with H, the angle the hori- 
zontal projection makes with GL, and the size of the pyramid 
being given. 

A careful examination of Figs. 32, 33, and 34 will be suffi- 
cient to understand this problem, since the principles are 
exactly the same as in the last problem. 

80. Prob. 15. To draw the projections of a prism 1" square 
and If" long, resting with one of its long edges on H, this edge 



94 PROJECTIONS. 

making an angle of 60° with V, backward and to the right, its 
front end being 2f " in front of V. The lower left-hand long 
face making an angle of 30° with H. 

Also, draw projections of a regular hexagonal pyramid whose 
altitude is 3", diameter of circumscribed circle about base is 1^-". 
The lower end of right-hand element of pyramid rests on H 1^" 
to the left of the point located in prism, and 1-J" in front ofY; 
this element also rests on top edge of prism at a point J" from its 
front end. The axis of pyramid slopes downward, backward, 
and to the left. The two lower faces of pyramid make equal 
angles with H. Fig. 62. 

The projections of the prism are drawn as already described 
in Art. 74 ; the spaces a h d h , d h b h , and b h c h are made respectively 
equal to ad, db t , and bc u . 

Heretofore, in drawing the projections of an object making 
a compound angle with the coordinate planes, we have had given 
the size of the object, the angle it made with V or H, and the 
angle its other projection made with GL, and the object has been 
drawn in three distinct positions. If the third position only is 
wanted it is not essential that the first two be wholly drawn, 
nor that they be made in separate figures. After the student 
becomes familiar with this work so that numerous lines do not 
confuse him, a considerable part of the construction may be 
omitted. In the figure all the necessary construction lines 
have been left in. 

Here neither the angle the pyramid makes with H, nor the 
angle its horizontal projection makes with GL, are given, but 
the projections of two points /and e of one of the elements are 
given, which enables the projections of this element to be drawn 
(indefinite in length). Reyolve this line around until it is parallel 
to V, and lay off on it in this position the true length of the ele- 
ment. This line will, of course, show the true size of the angle 



PROJECTIONS. 95 

this element makes with H, and the horizontal projection of 
the indefinite line before revolution shows the angle it makes 
with GL. All the necessary data are now obtained, and a care- 
ful study of the figure should enable the student to understand 
the rest of the construction. 

81. It is evident that neither the height of an object above 
H nor its distance in front of V affects in the least the size and 
shape of its projections. Therefore, the GL is not at all essen- 
tial in drawing the projections of an object unless its distances 
from V and H are given, which is not customary. In working 
drawings the GL is never used. It is only used in elementary 
projections as an aid in understanding the principles. 



CHAPTER V. 



SHADOWS. 



82. The shadow of a body upon a surface is that portion 
of the surface from which light is excluded by the body. 

The source of light may be assumed at any point, but it is 
customary to assume it so that the rays of light are parallel to 
the diagonal of a cube, as already stated in Ajt. 63, in which 
case its two projections make angles of 45° with GL. Al- 
though rays of light diverge in all possible directions from the 
source, yet when this source is as far removed as the sun there 
is no appreciable error in calling them all parallel. 

83. The shadow of a point on any surface is where a ray of 
light through that point pierces the surface. 

Hence, to find the shadow of a point on a surface, draw a 
line through the point to represent the ray of light, and find 
where it pierces the surface. 

84 To find the shadow of a point on H. 

In Fig. 35 let b represent the point in space and R the ray of 
light passing through this point. b v and U 1 will be the two pro- 
jections of the point b ; b v r and b h t of the ray of light. 

The shadow of the point b on H will be where R pierces H. 
This point being in H will have its vertical projection in GL, 
and its horizontal projection will be the point itself (Art. 33) ; 
being in the ray of light its two projections must also be on 
the projections of the ray of light (Art. 42-8th). Therefore, 
produce, if necessary, the vertical projection of the ray of light 



SHADOWS. 97 

till it meets GL at r, and r will be the vertical projection of the 
point where the ray of light pierces H ; at r draw a perpendic- 
ular to GL and JJ, where this perpendicular intersects the hori- 
zontal projection of the ray of light is its horizontal projec- 
tion, and is the shadow of the point b on the plane H. 
This is shown in actual projection in Fig. 37. 

85. To find the shadow of a 'point on V. 

In Fig. 36 let a represent the point and R the ray of light 
passing through it. a v and a h will be the projections of the 
point a; a v r and a h t of the ray of light. 

The shadow of the point a on V will be where R pierces V. 
This point being on V will have its horizontal projection in GL, 
and it will also be in the horizontal projection of R, hence at 
their intersection t; the vertical projection of this point, and 
the shadow on V, will be a v s where a perpendicular drawn from 
t to GL intersects d°r. 

This is shown in actual projection in Fig. 38. 

86. We have heretofore supposed that the coordinate planes 
did not extend below or behind their line of intersection, but 
they can just as well be considered as extending indefinitely in 
both directions, as shown pictorially in Figs. 35 and 36. Then, 
after the vertical plane has been revolved to coincide with the 
horizontal plane, that portion of the paper above GL repre- 
sents not only that portion of V which is above H, but that part 
of H which is behind V ; also that portion of the paper below 
GL represents that part of H which is in front of V and that 
part of V which is below H. 

Referring again to Fig. 35, we have already seen that R 
pierces H at ftj; now, if we suppose R to be produced below H 
indefinitely, it must pierce V at some point, since it is not par- 
allel to V. This point is found in exactly the same way as 
already described in Art. 85. It does not make a particle of 



98 SHADOWS. 

difference whether it pierces V above or below H. That is, 
every ray of light, unless parallel to V, will pierce V at some 
point either above or below H, and since these points are all 
in V their horizontal projections must be in GL. Of course, 
the shadow of the point b falls on H, and does not actually fall 
on V, but the point can be found where it would fall if H were 
transparent, and it is frequently convenient to do this in finding 
shadows of bodies in certain positions, as we shall soon see. 

Fig. 37 shows this point Q in actual projection. It being 
on that part of V which is below H after revolution appears 
below GL. 

Referring again to Fig. 36, it is evident that R not only 
pierces V at a v s , but also pierces H at a*. a* being in H has 
its vertical projection in GL. 

Fig. 38 shows this point in actual projection. It being on 
that part of H which is behind V, after revolution appears 
above GL. 

87. The following rules are evident from the foregoing : — 

To find the shadow of a point on H, produce the vertical pro- 
jection of the ray of light to meet GL ; erect a perpendicular at 
this point of intersection, and the intersection of this perpen- 
dicular with the horizontal projection of the ray of light will be 
the shadow required. 

It should be carefully borne in mind that this last perpendic- 
ular may intersect the horizontal projection of the ray of light 
above or below GL, depending on the location of the point in 
space ; that is, if the point be nearer H than V the intersection 
will be below GL, and the shadow will actually fall on H ; if 
the point be nearer V than H the intersection will be above GL, 
and the shadow will be imaginary. 

To find the shadow of a point on V, produce the horizontal 
projection of the ray of light to meet GL ; erect a perpendicular 



SHADOWS. 99 

at this point of intersection, and the intersection of this perpen- 
dicular with the vertical projection of the ray of light will be the 
shadow required. 

Here, also, the same caution as for the last rule is applicable. 
But in this case if the point is nearer V than H the intersec- 
tion is above GL, and the shadow actually falls on V ; if the 
point is nearer H than V the intersection will be below GL, 
and the shadow will be imaginary. 

The following may also be noted : — 

If the horizontal projection of the ray of light meets the GL 
before the vertical projection, the shadow will actually fall on V; 
if the vertical projection meets the GL before the horizontal, the 
shadow actually falls on H. 

88. Fig. 39 shows how to find the shadow of a line par- 
allel to both V and H. 

Fig. 40 shows the shadow of a line perpendicular to and 
resting on H. 

Fig. 41 shows the shadow of a line perpendicular to H but 
not resting on H. In this case a part of the shadow falls on 
each of the coordinate planes. 

From these figures the following facts may be noted : — 

1st. The shadoiv of a straight line on a plane surface is a 
straight line. 

2nd. The shadow of a line on a plane to which it is paral- 
lel is a line parallel and equal to it in length. 

3rd. The shadow of a line on a plane to which it is perpen- 
dicular coincides ivith the projection of the ray of light on that 
plane, and it is longer than the line itself. 

4th. The shadow of a line on a plane may be said to begin 
where the line pierces that plane, either or both being produced 
if necessary. 

5th. Since two points determine a straight line it is sufficient 



100 SHADOWS. 

to find the shadow of two points of it on a plane surface. In 
case the direction of the shadow, or the point where the line meets 
the plane surface receiving the shadow is known, it is sufficient 
to construct the shadow of one other point only. 

6th. When the shadow of a line falls upon two surfaces which 
intersect, the shadows on the two surfaces meet at a common point 
on their line of intersection. This is equally true whether the 
two surfaces intersect at right angles to each other or otherwise. 

89. Prob. 15. To find the shadow in V and H of a line 
oblique to V and H, when one end is nearer H than V, and the 
other is nearer V than H. Fig. 42.. 

Let a v b v and a h b h be the two projections of such a line. The 
end a being nearer H than V its shadow will fall on H (Art. 
87), and will be found, as already described, at a\. The end b, 
being nearer V than H, its shadow will fall on V, and will be 
found at b v s . Since the shadow of one end falls on H and of 
the other on V, it is evident that the shadow of the line will 
fall partly on H and partly on V, and also that a line joining 
these two points could only be a line in space, and therefore 
not the shadow required. 

It is essential that the two points which determine the shadow 
of a line should be on one and the same plane ; therefore, as we 
have the shadow of one point of the line on each coordinate 
plane, it is necessary to construct the shadow of another point 
of the line on either of the coordinate planes. Any point could 
be taken, but the ends being definitely projected it is more con- 
venient to use them. We have already found the shadow of a on 
H to be a% ; the shadow of b on H is found at b[f, in the same 
way. (This shadow we know does not actually fall on H, but 
it serves our purpose, which is to get the direction of the shadow 
of the line, just as well as if it did) ; a*b h s is, therefore, the 



Fig.26 



lFig.27 



1—^fe 





Fig. 2 8 



^ih \ / 




SHADOWS. 101 

shadow of the whole line on H, but only that part of it, a^c, 
which falls below or in front of GL is actual shadow. We 
have also seen that the shadow of b on V is b v s , the shadow of a 
on V is similarly found at <f s , and a 1 s b^ is the shadow of the 
whole line on V, but only that part of it, b s c, which falls above 
or behind GL is actual shadow. We, therefore, have as the 
actual shadow of the line a\c on H and b s c on V, the portions 
on the two planes intersecting in a common point, c, on GL as 
already noted in Art. 88-6th. Since this is the case it is evi- 
dent that it is not necessary to find the shadow of the whole 
line on both V and H ; having found the point c where either 
shadow crosses GL, join this point with the shadow of the end 
which falls on the other plane. 

90. Fig. 43 shows the shadow on V of a square card whose 
surface is parallel to V. Since the edges of the card are parallel 
to V their shadows will be parallel and equal each to each, and 
consequently the shadow of the card will be equal and parallel 
to the card itself. This will be true whatever the size or shape 
of the card. 

Therefore, the shadow of any plane figure on a surface to 
which it is parallel is a figure equal and parallel to it. 

91. Fig. 44 shows the shadow on V of a square card whose 
surface is perpendicular to V and parallel to H. Here the two 
edges ab and cd are parallel to V, consequently their shadows 
will be equal and parallel lines. The other two edges, ad and 
be, are perpendicular to V, hence their shadows make angles of 
45° with GL. 

92. Fig. 45 shows the shadow on both V and H of a square 
card whose surface is perpendicular to both V and H, that is, 
is in a profile plane. The edge cd is parallel to V, and is nearer 
to V than H, hence its shadow is on V parallel and equal to 
cd; the edge ad is nearer to V than H, and is perpendicular to 



102 SHADOWS. 

V, hence its shadow is on V, and makes an angle of 45° with 
GL ; the edge ab is parallel to V and perpendicular to H, the 
upper end is nearer V than H, and the lower end is nearer H 
than V, hence its shadow is partly on V and partly on H ; that 
portion which falls on V will be parallel to a v b v , and that por- 
tion which falls on H will make an angle of 45° with GL ; the 
other edge be is parallel to H and perpendicular to V, the front 
end is nearer H than V, and the back end is nearer V than H, 
hence its shadow will fall partly on Y and partly on H; that 
portion which is on H will be parallel to b h c h , and that portion 
which is on V will make an angle of 45° with GL. The 
shadow of the whole card on V is affigC^dg, of which only that 
portion a^nmc^d^, which is above GL, is visible. The shadow 
of the whole card on H is ajijcjrfj, of which only that portion 
b h s mn is visible. It is, of course, not essential to find that por- 
tion a h s nmc h s d h s of the shadow on H which is above GL. 

93. Fig. 46 shows the shadow of a card lying in a profile 
plane, all of its edges being oblique to both Y and H. Each 
point being found the same as all the preceding ones, no further 
explanation is necessary. 

94. Fig. 47 shows the shadow of a circular card parallel to 
H. The shadow on H we know must be a circle equal in size 
to the card, therefore it is only necessary to find the shadow of 
its centre Oi This is found at o\. With this point as a centre 
and a radius equal to that of the card describe the arc of a 
circle mm. Since a part of the circle is above GL it is evident 
that that part of the shadow actually falls on Y. To get this 
shadow take any points on the circle, as a,b,c,d, etc., and fiud 
their shadows separately, joining these points by a curved line. 
The points m and n, where the circle described from o h s as a centre 
crosses GL, will, of course, be two points on the curve. This 
curve will be an ellipse, and the shadow of a circle on a plane 
to which it is perpendicular or oblique will be an ellipse. 



SHADOWS. 103 

95. As a solid is composed of planes, planes of lines, and 
lines of points, it is evident that the shadow of the most com- 
plex body is obtained by finding the shadow of one point at a 
time by the methods already given until the shadows of all the 
points on the object which cast shadows have been found, so 
that the student who finds himself now able to cast the shadow 
of any single point on a given plane has practically mastered 
the subject, and if such a one has any difficulty in finding the 
shadow of any object the trouble is that he does not understand 
thoroughly the subject of projections. 

96. Since the shade lines of a body separate its light from 
its dark surfaces, the shadow of the shade lines will form the 
boundary of the shadow of the body. Therefore, in finding 
the shadow of a body the shade lines should first be marked, 
if it is in such a position that they can be found by means of 
the 45° triangle, and the shadows of these lines give the shadow 
of the whole object. If the object is in such a position that 
the shade lines cannot be found by means of the 45° triangle 
directly, the shadow of every point on the object, except those 
which it is known do not cast shadows, should be found sepa- 
rately, and then join those points which will enclose the largest 
area. The shade lines can then be found from the boundary 
of the shadow by finding what lines on the object cast these 
boundary lines. 

97. Prob. 16. To find the shadow of an hexagonal prism 
with its two ends parallel to H. Fig. 48. 

The shade lines of the prism in this position are found 
directly by means of the 45° triangle to be ab, be, cd, de, ef, 
fm, mn, and an. It is only necessary to find the shadow of 
each of these lines and join them in order, and the shadow is 
completed. The first three and last three of these lines are 



104 SHADOWS. 

parallel to H, hence their shadows on H will be equal and par- 
allel respectively to the edges casting them. 

98. Fig. 49 shows the shadow of a square prism on V and 
H, resting with its base on H and its long faces oblique to V. 
The shade lines are found first here. 

99. Fig. 50 shows the shadow of a cylinder on V and H, 
with its base resting on H. The shadow of any number of 
points on the shade line between a and d can be found. That 
portion of the curve between a v s and d v s will be a semi-ellipse, 
and the lines aim and d r s n must be tangent to this ellipse at the 
points a v s and d v s . 

100. Prob. 17. To find the shadow of a right cone resting 
with its base on H. Fig. 51. 

It is evident that, unless all the sloping part of the cone is 
light, there will be two elements of the cone which separate 
light from dark surfaces, and also that these two elements meet 
at the vertex and terminate at the other end in the base. But 
we do not know just where the light surface stops and the dark 
surface begins, as we do in the case of the cylinder, so we have 
to cast the shadow first. 

The shadow of the vertex o is o h s ; from o h s draw two lines 
o\a h and o\b h tangent to the base of the cone, and these will be 
the shadows of the two shade elements. We now see that oa 
and ob are the dividing lines between the light and dark sur- 
faces, but since these lines do not coincide with the contour 
elements od and oc, it is evident that neither od nor oc is a shade 
line. 

The bottom of the cone is dark and the sloping surface o h a h c h b h 
is light, hence the edge a h c h b h -is a shade line. 

Note the difference in the shade lines on the cylinder and 
the cone. 



SHADOWS. 105 

Do not shade the contour elements of a cylinder or cone. 

101. We have seen that to find the shadow of a body three 
things must be given, the body casting the shadow, the surface 
receiving the shadow, and the ray of light, and that these must 
be given by their projections. The surface receiving the shadow 
has for convenience so far been taken as one of the coordinate 
planes, one projection of which is the plane itself and the other 
is the ground line, but it is more frequently necessary to find 
shadows of objects upon themselves and upon other objects in 
the immediate vicinity. We have also seen that in finding 
shadows on the coordinate planes we have only concerned our- 
selves with that projection of the surface which is a line, that 
is, the GL. Hence, to find the shadow of an object on surfaces 
other than the coordinate planes, we have only to find the line 
which corresponds with GL and proceed according to the rule 
already given (Art. 87). 

To find this line, which we may call GL, observe the follow- 
ing rule : — 

Rule for GL. The GL to be used in finding shadows is 
always that projection of the surface receiving the shadow which 
is a line. This line may be straight, curved, or otherwise. 

Of course, this rule is applicable only when one projection of 
the surface is a line. 

102. Prob. 18. To find the shadow of a stick of timber on 
the top and front of an abutment on which it rests. Fig. 52. 

The shade lines of the stick of timber are easily found to be 
as follows : cd, dd, de, and efi First find the shadow on the 
top surface of the abutment ; the vertical projection of this sur- 
face is the line a v b v , which, according to the rule, is the GL to 
be used. The shadow of dd is d h d h s , of de is d h s e h s , and of ef 
(since efis parallel to the top of the abutment) is a line through 



106 SHADOWS. 

e\ parallel to e 7 f h . Next find the shadow on the front surface 
of the abutment; the horizontal projection of this surface is 
the line a h b h , which must therefore be the GL for this surface. 

The shadow of the shade edge ef will fall on both the top 
and front surfaces, which intersect in the line ab, therefore the 
shadows on the two surfaces will meet in a common point m on 
this line (Art. 88-6th). m v is, then, one point of the shadow 
of the edge ef on the front of the abutment. The point n v , for 
the same reason, is one point of the shadow of the edge cd t on 
this surface. Since these two edges are parallel they will cast 
parallel shadows, hence it is only necessary to find the shadow 
of one more point, and the shadow is determined. The shadow 
of the point c is c v s ; join this with n v , and through m v draw a 
line parallel to c v s n v , and the shadow is completed. Of course, 
any other point might have been taken on the edge cd t , or any 
point on the edge ef 

The irregular line r h s h simply indicates that the abutment 
extends backward farther than it was necessary to show. The 
ragged end of the stick of timber indicates the same thing. 

103. Prob. 19. To find the shadoiv of a stick of timber 
on another stick of timber into which it is framed. Fig. 53. 

It is evident that the sloping stick will cast a shadow on the 
top of the horizontal stick. The lower front and the upper back 
edges cd and ab are the lines which cast the shadows, and mn is 
the line to be used as GL. The shadow of the point a is at 
aj, apparently on the elevation of the object, but in reality it is 
where the shadow would be if the lower stick of timber were 
sufficiently wide to receive it. It is frequently necessary to 
imagine surfaces indefinite in extent for convenience in con- 
struction. The shadow of the edge ab begins where it leaves 
the surface (Art. 88-4th), therefore, join a h and b h and that 



SHADOWS. 107 

portion of the line which falls on the surface of the stick will 
be all that is necessary ; through c h draw a line parallel to b h a h , 
and the shadow is completed. 

104. Prob. 20. To find the shadow of one oblique stick of 
timber on another, both being parallel to V and lying against each 
other. Fig. 54. 

The shadow will fall on the front and top of the back stick. 
The ground line for finding the shadow on the front surface is 
rs, and for the top surface is mn. Here, as in Prob. 18, the 
horizontal projections of the points c° and d v , where the shadow 
leaves the front surface, will be points of the shadow on the top. 
A careful examination of the figure will make further explana- 
tion unnecessary. 

105. Prob. 21. To Jind the shadow of a straight wire 
lying on top of a vertical cylindrical wall on the wall and the 
shadow of the wall on itself Fig. 55. 

Here the GL is a curved line, and is to be used just the same 
as heretofore. There being no new principles the student should 
be able to understand this problem from an examination of the 
figure. 

106. Prob. 22. To find the shadow of the head of a bolt 
on its shank when the length of the bolt is parallel to both V 
and H. 

In Fig. 56 the head is hexagonal and the shank is cylindri- 
cal, and in Fig. 57 the head is octagonal and the shank is 
hexagonal. 

In cases like these it is evident that neither the plan nor ele- 
vation of the surfaces receiving the shadow is a line, but the end 
view, or the projection on a profile plane, is a line, and accord- 



108 SHADOWS. 

ing to the rule is, therefore, the GL to be used, mno is the GL 
in Fig. 56, and mnor in Fig. 57. The only other new point to 
be noticed is that the two elevations of the ray of light each 
make an angle of 45° with a horizontal line, and slope in the 
same direction. 

107. Prob. 23. To find the shadow of a chimney or tower 
of a house on the roof Fig. 58. 

The end view of the roof mn is the GrL for this problem. 
The shadow is found the same as in Figs. 56 and 57. 

108. Prob. 24. To find the shadow of a stick of timber on 
the top and sloping faces of an oblique abutment. Fig. 59. 

Since neither plan, elevation, nor end view of these sloping 
surfaces is a line our rule is no longer directly applicable, and 
we may make use of an indirect method (the shadow can be 
found directly by descriptive geometry methods). 

First find the shadow on top of the abutment. The points 
c h and e\ where it leaves the top, will be two points of the 
shadow on the left, front, sloping face (Art. 88-6th), c° and e v 
will be their vertical projections ; in the same way d h and n h 
will be two points of the shadow on the right, front, sloping face, 
and d v and ri° will be their vertical projections. The shadow 
of the lower front edge (the upper back edge could just as well 
have been taken) on the ground is the indefinite \\\\Qf h o h . The 
ground and the sloping face of the abutment intersect, hence 
the pointy 7 * will be a point of the shadow on the left front face ; 
f v will be its vertical projection ; for the same reason o h is a 
point of the shadow on the right, front face, and o v is its verti- 
cal projection. Join e h f h , and through c h draw a line parallel 
to it; also, n h o h and a line through d h parallel to it. This com- 
pletes the shadow on the plan. The shadow on the elevation 
is found by projection from the plan. 



SHADOWS. 109 

If the abutment had been so high that the shadow on the 
ground would have been difficult to obtain, an imaginary hori- 
zontal plane could have been taken at any convenient place, 
and the shadow found on that, noting where it comes out from 
the abutment. The dash and two dots line in plan and eleva- 
tion represents the two projections of the intersection of such 
an imaginary plane with the abutment. m h t h is the shadow of 
the lower front edge of the stick on this imaginary plane, which 
gives the points t h and m h as points of the shadow. The shadow 
is completed as before. 

109. Prob. 25. To find the shadow of any oblique line on 
any oblique plane. Fig. 60. 

First find the shadow on any horizontal auxiliary plane. 
m v e v will be the vertical projection of one such plane, and 
will be the GL for this plane. The shadow of any point as a 
on this plane will be a* ; the shadow of any other point could 
be found in the same way, thus getting the shadow of the whole 
line, but the shadow of a line begins on a plane where the line 
pierces the plane, and the line pierces this auxiliary plane at 
the point b (b v being where m v e v intersects the vertical projec- 
tion of the line, and V 1 being perpendicularly below it on the 
horizontal projection of the line), therefore, joining a h s and V 1 
we have the shadow on the auxiliary plane. m v e v is the vertical 
projection of the line of intersection of the auxiliary plane with 
the card mnop ; its horizontal projection is, therefore, m h e h . The 
lines b h a h s and m h e h lie in the auxiliary plane, the line m h e h also 
lies in the plane of the card mnop, therefore, the point r\ where 
these two lines intersect, is a point on the card, and must be one 
point of the horizontal projection of the shadow required ; r v on 
m v e v is its vertical projection. Assume any other auxiliary 
plane, as d v o v , and another point, s v s h of the shadow, will be 



110 SHADOWS. 

found in the same way. Draw an indefinite line through these 
points r and s in both projections, and the shadow is finished. 

Vertical auxiliary planes could have been taken instead of 
horizontal with the same result. 

110. Prob. 26. To cast the shadow of an abacus on a con- 
ical column. Fig. 61. 

Since neither projection of the surface receiving the shadow 
is a line, this problem must be done by the indirect method as 
used in the two preceding problems. 

Find the shadow on any horizontal plane, as a v c v . Its ver- 
tical projection is a v c v , and is the GL for this shadow. This 
plane cuts the column in a horizontal circle, of which a v c v is 
the vertical and a h b h c h is the horizontal projection. The shadow 
of the bottom edge of the abacus on this plane is the circle 
d h a h e h , drawn with oJ, the shadow of o the centre of the abacus, 
as a centre and radius equal to that of the abacus. This circle 
cuts the circle a h b h c h at the point a h , which is one point of the 
horizontal projection of the shadow required. a v , on the verti- 
cal projection of this auxiliary horizontal plane, is one point of 
the vertical projection of the required shadow. Any number 
of other points can be found in the same way. 

111. Prob. 27. To cast the shadow of the prism given as 
in Fig. 62 on H and V, also of the pyramid on the prism and 
on H. 

The shadows of the prism on H and V and of the pyramid 
on H require no explanation. 

If the shadow of a line falls partly on two plane surfaces, 
A and B (no figure), which do not intersect (at that portion 
under discussion), A being between the line and B, that part 
which falls on A does not fall on B, and the shadow of the line 



SHADOWS. Ill 

on B may be said to begin at the shadow of the point where 
the shadow of the line leaves A, on B. Now, if B is horizontal 
or vertical, and A is oblique to both V and H, the shadows of 
the line and the plane A on B can be readily found ; and then 
get the shadow of the line on A by the reverse of the above 
process, that is, note where the shadow of the line and plane A 
on B intersect; find what point on A cast this intersecting 
point, and that will be one point of the shadow of the line on A. 

Now, referring to Fig. 62, we see that the shadow of the 
pyramid and prism on H intersect at the four points rj, *J, a£, 
and zj ; the points on the prism (this being between the pyra- 
mid and H) which cast these shadows are r h , t h , x h , and z h 
respectively (found by drawing a 45° line from the shadow to 
the edge casting it), and they are, therefore, points of the 
shadow of the pyramid on the prism. 

The shadow on the upper edge of the prism may be found 
by this same principle, that is, the shadow of this upper edge 
on H is b h s y h s ; this intersects the shadow of the pyramid on H 
at s h s and yj ; s h and y h are the points which cast these shad- 
ows, hence they are points of the shadow required. Join rV*, 
s h t h , x h y h , and y h z h , and the horizontal projection of the shadow 
is completed. Its vertical projection is found by projecting 
each one of these points on to the vertical projection of the 
prism. The points s h and if 1 could also have been found by 
casting the shadow of the pyramid on an auxiliary horizontal 
plane through the top edge of the prism ; o %s h and o*y h are the 
shadows of the two shade elements of the pyramid on such a 
plane ; they intersect the upper edge of the prism at the points 
s h and y h , which are the points required. A vertical auxiliary 
plane through this edge or the other edges of the prism might 
have been used with the same result. 



CHAPTER VL 



ISOMETRICAL DRAWING. 



112. In all the previous constructions two projections have 
been used to represent a body in space. In isometrical projec- 
tions only one view is used, the body being placed in such a 
position that its principal lines or edges (length, breadth, and 
thickness) are parallel to three rectangular axes, which are so 
placed that equal lengths on them are projected on the plane 
equal to each other. Thus we have the three dimensions of a 
body shown on one plane in such a way that each can be meas- 
ured, thereby combining the exactness of ordinary projections 
and the intelligibleness of pictorial figures. It is used chiefly 
to represent small objects in which the principal lines are afc 
right angles to each other. In large objects the drawing would 
look distorted. 

113. If we take a cube situated as in Fig. 63, and tip it to 
the left, about its lower left corner e until the diagonal eg is 
horizontal, Fig. 64, and then turn it through an angle of 90°, 
still keeping eg horizontal, we obtain Fig. 65. The vertical 
projection in this figure is what is called an isometrical projec- 
tion. 

The edges of a cube are all of equal length, and it will be 
seen that they appear equal in the figure, consequently the visi- 
ble faces must appear equal. It will also be seen that the figure 
can be inscribed in a circle, and that the outline of the isomet- 
rical projection is a regular hexagon, hence, that those lines 



ISOMETRICAL DRAWING. 113 

which represent length and breadth make angles of 30° with 
a horizontal, and those which represent thickness are vertical. 

114. The edges of the cube being inclined to the plane on 
which they are represented, appear shorter than they actually 
are on the object, but since they are all equally foreshortened, 
and since a drawing may be made to any scale, it is customary 
to ignore this foreshortening, and make all the isometrical lines 
of the object equal to their true lengths. This will give what 
is called the isometrical drawing of the object, which will be 
somewhat larger than the isometrical projection. 

Fig. 65 represents the isometrical projection of the cube 
shown in Fig. 63, and Fig. 67 is the isometrical drawing of the 
same cube. 

115. Definitions. c v , Fig. 65, is the isometric centre. 
c v b v , c v d v , and c v e v are the isometric axes. Lines parallel to either 
of the isometric axes are isometric lines, while any line not 
parallel to one of these axes is a non-isometric line. Planes 
parallel to the faces of the cube are isometric planes, and those 
which are not parallel to one of these faces are non-isometric 
planes. -^ 

116. Direction of Light. In isometrical drawing the 
light is supposed to be parallel to a diagonal of the cube, as in 
ordinary projections, only here it is parallel to the plane of 
projection, and it is represented by a line making an angle of 
30° with a horizontal line. Any line parallel to df, Fig. 67, 
may represent a ray of light. 

117. Shade Lines. These are the same as in ordinary 
projections, that is, they separate light from dark surfaces. In 
all rectangular objects the top, left front, and left back sur- 



114 ISOMETRICAL DRAWING. 

faces are light, while the bottom, right front, and right back 
surfaces are dark. In Fig. 67 the edges ab, be, ce, and el are 
the visible shade lines. And all rectangular objects have their 
shade lines in relative positions to those of the cube. As in 
ordinary projections, in putting shade lines on a group of ob- 
jects touching each other, the group is shaded as if it were one 
solid ; also, in outline drawing the shadows are disregarded in 
putting in shade lines. 

118. Prob. 28. To make the isometrical drawing of a cube. 
Fig. 67. 

With the centre c and radius equal to the edge of the cube 
draw a circle, and in it inscribe a regular hexagon, and draw the 
alternate radii cb, cd, and ce, and the drawing is completed. 

Another method, which is applicable to any rectangular 
object, is to draw from any point as e lines e/and el, each mak- 
ing an angle of 30° with a horizontal, and the vertical line ce ; 
on these lay off ef, el, and ec equal to the true length of the edge 
of the cube ; from the points/* and I draw indefinite vertical lines ; 
from c draw the lines cb and cd parallel to ef and el, intersect- 
ing the verticals through f and I in the points b and d; from 
the points b and d draw lines parallel to cd and cb, meeting in 
a. This completes the drawing. 

119. Prob. 29. To make the isometrical drawing of a rect- 
angular block, with another rectangular block resting on its top 
face, and a recess in its right front face. Fig. 6Q. 

Construct the isometrical drawing of the large block by the 
second method of Art. 118. 

To draw the small block it is first necessary to locate one of 
its lower corners in the top face of the large one. This must 
be done by means of two coordinates referred to two isometric 



ISOMETRICAL DRAWING. 115 

lines as axes. The point e, in the top face of the large block, 
is -J" from the side and T 3 F " from the end, therefore make of 
equal to -J" and ag equal to T V ? then from f and g draw the 
isometric lines ef and ge, intersecting in e, the point required. 
The rest of the small block is drawn in the same way as the 
large one. 

To make the recess in the front side the point t is located 
in the same way as the point e, and tx, equal to the depth of 
the recess, is laid off as shown. The rest of the construction 
is evident. 

120. Fig. 68 shows the isometrical drawing of a rectangular 
box, without cover, 15" long, 6" wide, and 4" high, outside 
measurements, the boards being f" thick. The scale being 
1^-" = 1'. The ends are nailed on to the sides, and the bottom 
is nailed to the sides and ends. The visible joints are shown. 
The dotted lines show the inner edges of the box which are 
not visible. 

121. Fig. 69 shows the isometrical drawing of a four-armed 
cross. A careful examination of the figure will enable the 
student to understand its construction, there being only iso- 
metric lines involved. 

122. Prob. 30. To make the isometrical drawing of the 
pentagonal prism shown in Fig. 20. Fig. 70. 

The edges of the base, not being at right angles to each other, 
are non-isometric lines, hence the base should first be inscribed 
in a rectangle. Let one side of the rectangle contain the edge 
fo, and the other sides respectively contain the corners a, d, 
and e. Make the isometrical drawing of the rectangle and 
locate each corner of the base, a, b, c, d, and e, by laying off on 
the sides of this rectangle the distance of each point respect- 



116 ISOMETRICAL DRAWING. 

ively from the nearest corner of the circumscribed rectangle. 
At these points draw vertical lines, and lay off on each of them 
the true height of the prism and join the tops, completing the 
drawing of the prism. It will be noticed that the non-isometric 
lines on the drawing are not equal to their true length on the 
object. 

123. Prob. 31. To make the isometrical drawing of an 
oblique timber framed into a horizontal one, as given in Fig. 53. 
Fig. 71. 

The horizontal timber is drawn as usual. To draw the 
oblique one, these edges being non-isometric, two points have 
to be located by means of coordinates. The point a is located 
by making ad equal to the distance the lower end of the oblique 
stick is from the end of the horizontal one. Any other point 
b is found by making cd equal to the horizontal distance of the 
point b from d, and be equal to its vertical distance from d. Join 
ab, which gives the isometrical drawing of one edge of the ob- 
lique timber. The other edges are, of course, parallel to this, 
being drawn through the points e and f which are located the 
same as point e in Prob. 30. 

124. Prob. 32. To make the isometrical drawing of the 
skeleton frame of a box made in the form of the frustum of a 
square pyramid. Fig. 72. 

Let a square prism be circumscribed about the frustum. 
The isometric of this prism is readily drawn, and is shown by 
the dotted lines. The bottom edge of the frustum coincides 
with the bottom of the prism. The points a, b, c, and d are in 
the upper face of the prism, and are found as the point e is in 
Prob. 30. 

Join ab, be, cd, da, af bg, and de, and the main frustum is 



ISOMETEICAL DRAWING. 117 

completed. The other lines which change the frustum from a 
solid to a skeleton need no explanation. 

125. Prob. 33. To make an exact isometrical drawing of 
a circular card, and also of any scroll or letter on its surface. 
Fig. 74. 

Let the circle and letter G be given as in Fig. 73. First, 
circumscribe a square about the circle. Make the isometric 
drawing of the square. The centres of the sides of the square, 
d, e, f and g, give four points of the isometric of the circle. 
Each point on the circumference of the circle, as a, b, c, etc., 
has two coordinates, by means of which the isometrical draw- 
ing of the points may be easily obtained. There will be four 
points on the circle whose coordinates will be the same as those 
for c, and eight points whose coordinates will be the same as 
those for a or b, or any point between d and e. The more 
points that are taken the more accurate will be the ellipse which 
forms the isometrical drawing of a circle. 

The letter is drawn in the same way by taking the two coor- 
dinates of any point on it. 

126. Prob. 34. To make an approximate construction of 
the isometrical drawing of a circle. Fig. 75. 

Make the isometrical drawing of the circumscribed square as 
before ; d, e, f and g will be four points. Draw the lines ag, 
of, bd, and be, intersecting in the points c and o ; c will be the 
centre of the arc between d and g ; o of that between e andjf; 
a of that between g and/; and b of that between d and e. 

127. Fig. 76 shows the approximate construction of the 
isometrical drawing of circles in each of the three visible faces 
of a cube. No explanation is necessary. 



118 ISOMETRIC AL DRAWING. 

The student should study this and Fig. 75, so as to be able 
to make the isometrical drawing of a quarter of a circle in 
either isometric plane without making the whole circle. 

128. Fig. 78 shows the isometrical drawing of a bolt, an 
hexagonal nut, and a circular washer as shown in Fig. 77. 

129. Fig. 79 shows the isometrical drawing of the hollow 
cylinder given as in Fig. 27. 

130. Prob. 35. To divide the isometrical drawing of a 
circle into equal parts. Fig. 74. 

At the middle point f of one of the sides nl of the isomet- 
rical drawing of the circumscribed square draw fo perpendicu- 
lar to nl, and make fo equal to the radius of the circle ; draw 
ol and on ; with o as a centre and radius fo describe the arc 
rfs ; divide this arc into any number of parts ; draw through o 
and these points of division lines to meet nl ; join these meet- 
ing points with t, the centre of the ellipse, and where these lines 
intersect the ellipse will be the points of division of the iso- 
metrical drawing of one quarter of the circle. 

The other quadrants can be divided in the same way. 

Another Method. On the long diameter of the ellipse 
draw the semicircle ckp ; divide this into any number of equal 
parts ; through these points of division draw lines perpendicu- 
lar to cp ; where these lines intersect the ellipse will be the 
points of division sought. 

131. In isometrical drawing the shadow of a point on a plane 
is where the ray of light through the point intersects its projec- 
tion on that plane. 

To find the shadow of the line dn on the top of the cube. 
Fig. 67. 

A ray of light through the point n is nn s , its projection on 



ISOMETRICAL DRAWING. 119 

the top of the cube is dn s ; these two lines intersect at the point 
n 8 , which is the shadow of the poyit n on the top of the cube. 
Join this point with d, and dn s is the shadow of the vertical line 
dn required. 

To find the shadow of the line do on the left, front face of the 
cube. 

A ray of light through the point o is oo $ , its projection on the 
face of the cube is do s ; these lines intersect at the point o s , 
which is the shadow of the point on the left, front face of the 
cube. Join this with d, and do s is the shadow of the line required. 

132. Prob. 36. To find the shadow of a cube on the plane 
of its base. Fig. 67. 

The shadow of the edge ec is ec s ; of the back edge ac is 
ca s ; of the point b is b s ; therefore, the shadows of ab and be 
are a s b s and b s c s respectively, which completes the shadow re- 
quired. 

133. In Figs. 71 and 78 the shadows cast by the objects on 
each other and on the ground are shown. 

134. To find the shadow of any point on any horizontal 
isometric plane proceed as follows : Draw though the point a 
vertical line, and make it equal in length (downward from the 
point) to the height of the point above the plane receiving the 
shadow ; through the upper end of this line draw a line at 30° 
to a horizontal in the direction of the ray of light ; through the 
lower end draw a horizontal line ; where these two last lines 
intersect will be the shadow of the point. 

To find the shadow of a point on any vertical isometric 
plane : Draw through the point a line at 30° to a horizontal, 
backward and to the right, and make it equal in length (back- 
ward from the point) to the perpendicular distance of the point 



120 I&OMETRICAL DRAWING. 

from the plane ; through the front end of this line draw a line 
parallel to the ray of light ; through the back end draw a line 
at 60° to a horizontal, forward and to the right ; where these 
two last lines intersect will be the shadow of the point. 

OBLIQUE PROJECTIONS. 

135. Oblique projections differ from isometrical projections 
only in the position of the principal faces of the object. In- 
stead of being placed so that its principal faces make equal 
angles with the plane on which it is represented, one of them 
is placed parallel to this plane, while the edges which represent 
the remaining dimension of the object may be drawn at any 
angle with a horizontal, for convenience usually at 30° or 45°. 
With this difference, all the statements and principles of iso- 
metrical drawing are equally applicable to oblique projections. 

136. Fig. 80 shows the oblique projection of a cube, the 
shadow of lines on its top and front faces, the shadow of the 
cube on the plane of its base, and the manner of drawing the 
oblique projection of circles in the three faces of the cube. 
The ellipse in the top face is drawn by the arcs of circles, 
and that in the right face by points located exactly by coordi- 
nates. 

It will be noticed that the ray of light is parallel to the diag- 
onal df of the cube, and that its projection on a principal sur- 
face is parallel to the diagonal db or de. 

137. Fig. 81 shows the oblique projection of the hollow 
cylinder given as in Fig. 27, of which Fig. 79 is the isometri- 
cal drawing. Figs. 1, 35, and 36 are oblique projections of 
models representing the principles of projections and shadows. 



Fig.69 



Fig. 73 



Fig. 72 




CHAPTER VII. 



WORKING DRAWINGS. 



138. A working drawing is one which shows all the dimen- 
sions of an object in such a way that the object could be repro- 
duced or constructed from the drawing. Two views at least, 
plan or horizontal projection, and elevation or vertical projec- 
tion, are necessary ; but more frequently a third view, usually 
an end view, is also necessary. Besides these views, one or 
more sections depending upon the object are sometimes neces- 
sary to completely determine all of its dimensions. 

139. If an object is cut through, in any direction, by an 
imaginary plane, the projection of one part of the object on a 
plane parallel to the cutting plane, the person supposed to be 
facing this cutting plane, is called a section. Sections are taken 
to show the form and dimensions of the interior of a hollow 
object, and also of some parts of solid objects which are not 
completely determined by its plan and elevations. These im- 
aginary cutting planes are usually taken either vertical or hori- 
zontal, but it is sometimes necessary to take them in other 
positions, but perpendicular to a vertical or horizontal plane. 
All that part of the object which is cut by these imaginary 
planes is cross-hatched, that is, is covered by parallel lines quite 
near together. The direction of these lines should be either 
30°, 45°, or 60°, and from ^\j" to -J" apart, depending on the size 
of the surface to be cross-hatched, the smaller the nearer to- 
gether. They may be drawn in either direction. 



122 WORKING DRAWINGS. 

140. Fig. 82 represents the elevation, Fig. 83 the plan, and 
Fig. 84 a vertical section through the centre on the line AB, 
of a stuffing-box gland. The side elevation was not necessary 
in this case, as all the dimensions of the solid are shown with- 
out it. These figures are made one-half size. 

141. It is not sufficient simply to draw the projections of the 
object the correct size, but the dimensions of the solid should 
all be clearly placed on the drawing, so that the workman, or 
whoever has occasion to read the drawing, is not obliged to 
use the scale, thereby removing a great liability to error. These 
dimensions should be put in neatly, and should follow a certain 
system. The method used in Figs. 82, 83, and 84 should be 
carefully observed, besides which the following general direc- 
tions for putting in dimensions and representing special feat- 
ures should be followed : 

In placing dimensions upon the drawing a line should be 
drawn from one point to another, between which the dimension 
is to be given, and the actual dimension, or distance apart of 
the points, is placed in the line, a space having been left for it 
near the centre. These lines should be fine and composed of 
dashes about ■§•" long with about -J" spaces between them. When 
the dimension is small of course the dashes must be made 
shorter. Arrow heads are placed at the ends of these lines, 
the point of the arrow exactly touching the points or lines be- 
tween which the dimension is given, the arrow heads pointing 
away from each other. When the dimension is very small the 
arrow heads may be placed on the outside of the lines instead 
of between them, and in that case should point toward each 
other. 

The arrow heads should be drawn free-hand and not made 
with the drawing pen. The figures for dimensions should also 
be made free-hand, and should always be placed at right angles 



WORKING DRAWINGS. 123 

to the dimension line, and should read from the bottom or right- 
hand side of the drawing. They should be put down in inches 
and thirty-seconds, sixteenths, eights, quarters, and halves, as 
the case may be, thus, l T y, not f £". The fractions should be 
reduced to their lowest terms, thus, f", not -{%". The dividing 
line in the fraction should always be made horizontal, as they 
are less likely to be misunderstood. The inch marks should 
be placed after the fraction. 

On rough castings measure to the nearest sixteenth, on ordi- 
nary finished surfaces take the nearest thirty-second, and on 
fine finish and fits be as accurate as possible. 

All dimensions up to two feet should be put down in inches, 
thus, 15", 22f", and all above that in feet and inches, thus, 
3'-2", or 3 ft. 2". Students should be very careful to get all 
of the important dimensions on, and also an " overall " dimen- 
sion, so that the workman will not be required to add a num- 
ber of dimensions together. Important dimensions are those 
which are necessary for the workman to construct the piece. 
The dimensions should not interfere with each other, and care 
should be taken not to have them cross each other in a circle. 

As a general thing do not repeat any dimensions ; that is, if 
a dimension is given on one view do not repeat it on another. 

In order that the drawing may be left as distinct as possible 
it is frequently advisable to put the dimensions outside the fig- 
ure, or better, where two or more views are given, put them 
between the different views, as shown in Figs. 82, 83, and 84 ; 
the widths being placed between the plan and elevation, and 
the heights between the elevation and section, or between the 
two elevations where an end elevation is given. To do this 
" extension " lines must be used. They should be composed of 
fine dash lines, the dashes being about -J" long, so as to be dis- 
tinguished easily from the dotted lines representing the invisi- 



124 WORKING DRAWINGS. 

ble parts of an object. Where the dimensions do not interfere 
with the drawing, as is the case in Fig. 84, it is better to put 
them on the figure between the lines themselves, or as near as 
possible. 

Give the diameter of a circle instead of the radius. When 
only an arc is shown give the radius, and draw a very small 
circle about its centre, and let this circle take the place of an 
arrow head. The dimension line should be drawn from the 
edge of this circle and not from its centre. 

In locating holes or bolts the dimensions should be given 
from the outside of the piece to the centre of the hole or bolt, 
and their distance apart is shown by giving the distance be- 
tween centres. See Fig. 85. Holes are very often located 
from the centre line of a piece, so that it is unnecessary to give 
the dimension from the outside of the piece. 

The " centre " line should be composed of long and short 
dashes, the long dashes being about J-" long and the short ones 
about £" long. 

If the holes are arranged in a circle, as in Fig. 86, give the 
diameter of the circle passing through the centre of the holes. 

In drawing a bolt or screw represent the threads as shown in 
Fig. 87 ; it is not necessary that the spaces should correspond 
with the true pitch of the threads. In order to obtain the cor- 
rect slant of the threads, a line drawn at right angles to the 
axis of the bolt should pass through the point of a thread on 
one side, and the centre of a space on the other. 

Always give that view of a square, hexagonal, or octagonal 
bolt-head, or nut which shows the distance between its paral- 
lel sides. 

In placing the dimensions upon a bolt or screw, always give 
the length of the unthreaded part in addition to the length of 
the bolt. The length of the bolt should be given from the 
under side of the head to the extreme end. See Fig. 87. 



Fig. 8 2 



M'h Fj g- 84 



Fig.92 



*5r~7r 



W 



W 1 \& Q M 



Fig.83 







Fig.85 




Fig. 90 



WOEKING DEAWINGS. 125 

In making a sectional view on a line passing lengthwise 
through the centre of a shaft, bolt, or screw, it is generally 
unnecessary to represent the shaft, bolt, or screw in section, as 
the view is more clearly shown by leaving them in full. See 
Fig. 89. 

In drawings where bolts or screws are shown by dotted lines 
do not dot in the threads, but represent them by double dotted 
lines, as shown in Fig. 89. 

Represent a tapped hole as shown in Fig. 90. 

The line on which a section is taken, as AB, Fig. 83, should 
be made the same as a centre line. 

142. Although it is customary to represent a screw by 
straight lines, as shown in Fig. 87, it is sometimes desirable to 
make its actual projections, especially if the screw be a large 
one. 

The thread of a screw is a curve which is called a helix. A 
cylindrical helix is generated by a point caused to travel round 
a cylinder, having, at the same time, a motion in the direction 
of the length of the cylinder, — this longitudinal motion bear- 
ing some regular prescribed proportion to the circular or angu- 
lar motion. The distance between any two points which are 
nearest to each other, and in the same straight line parallel to 
the axis of the cylinder, is called the pitch, — in other words, 
the longitudinal distance traversed by the generating point dur- 
ing one revolution. 

To draw the projections of a helix. Fig. 91. 

The plan of the helix will be a circle. Divide this circle 
into any number of equal parts, in this case twelve ; divide the 
pitch into the same number of equal parts. It is evident that 
when the point has moved one-twelfth the distance around the 
circumference, it has also moved in the direction of the axis 
one-twelfth of the pitch ; when it has moved two-twelfths the 



1 26 WORKING DRAWINGS. 

distance around it has moved two-twelfths of the pitch ; there- 
fore, from the points of division, a h , b h , c\ etc, in the plan draw 
vertical lines until they intersect horizontal lines drawn from 
the corresponding division of the pitch. And these intersec- 
tions, a v , b v , c v , etc., will be points on the vertical projection of 
the helix. 

Fig. 92 shows a V-threaded screw in projection. 



CHAPTER VIII. 



EXAMPLES. 



All polygons referred to in these Examples are regular polygons, unless other* 
wise stated. 

I. Draw the two projections of a point 1-J-" from H and l" 
from V. 

2* Of a point lying in H and f " in front of V. 

3. Of a point lying in V and l" above H. 

4. Of a line 1" long, parallel to both V and H, f" above H 
and 1" in front of V. 

5. Of same line when it is perpendicular to V and 1^" 
above H, its back end being ^" in front of V. 

6. Of same line when it is perpendicular to H and 1" in 
front of V, its lower end being J /# above H. 

7. Of same line when it is parallel to H, 1" above H and 
making an angle of 30° with V, its back end being J" in front 
of V. 

8. Of same line when it is parallel to V, f " in front of V 
and making an angle of 60° with H, its lower end being £-" 
above H. 

9. Of same line lying in H and making an angle of 45° 
with V, its back end being -J" in front of V. 

10. Of same line lying in V, parallel to and 1" above H. 

II. Of same line when it is inclined at an angle of 60° with 
H, and whose horizontal projection makes an angle of 45° with 
GL, one end being J" above H and J-" in front of V. 



128 EXAMPLES. 

12. A wire 1-|" long projects from a vertical wall at 60° 
with the surface, and is parallel to the ground and 1" above it. 
Draw plan and elevation. 

13. Find true length of a line given by its projections, as 
follows : One end is ^" from each plane and the other is 2" 
above H, the horizontal projection of the line is 1-J" long and 
makes an angle of 30° with GL. 

14. Draw plan and elevation of a line lj-" long, lying in a 
profile plane, making an angle of 60° with H, whose lower end 
is i" from V and J" from H. 

15. Of an oblique line, one end being above and in front of 
the other. Find its true length and angle it makes with H. 

16. Of an oblique line, one end being behind and above 
the other. Find its true length and angle it makes with V. 

17. Of a line which slopes downward, backward, and to 
the right. Find its true length by revolving parallel to V. 

18. Of a line which slopes downward, forward, and to the 
left. Find its true length by revolving parallel to H. 

19. Of two lines which are parallel in space and slope 
downward, forward, and to the right. 

20. Of a line 2" long, sloping downward, forward, and to 
the right, one end being 1-|-" above H and ^" in front of V, the 
other end -J" above H and 1^" in front of V. 

21. Draw plan and elevation of a rectangular card f" x 1 J" 
which is perpendicular to H, parallel to V, and f " in front of 
V ; its short sides are parallel to H and the lower one is J" 
above H. Revolve this card forward about its left-hand edge 
(like a door on its hinges) through angles of 30°, 45°, 60°, 
and 90°, and construct corresponding plans and elevations. 

22. Of same card when it is lying on H with its long sides 
parallel to V and J" in front of V. Revolve card about right- 
hand horizontal edge (like a trap-door on its hinges) through 



EXAMPLES. 129 

angles of 30°, 45°, 60°, and 90°, and construct corresponding 
projections. 

23. Of same card when, besides making the angles of 0°, 
30°, 45°, 60°, and 90° with H, as in last example, the hori- 
zontal projection of its long edges, in all its different positions, 
makes an angle of 30° with GL backward and to the right. 

24. Of same card when it is parallel to V and J" in front 
of V, one of its diagonals being parallel to H. Revolve this 
card through an angle of 60° about a vertical axis, and construct 
its corresponding projections. 

25. Of same card when, besides making an angle of 60° 
with V, as in last example, the vertical projection of the diago- 
nal which was parallel to H makes an angle of 45° with GL. 

26. Of a card J" square resting on H, with one diagonal 
parallel to V and 1" in front of V, then raise left-hand end of 
diagonal until it makes an angle of 45° with H and construct 
corresponding projections. 

27. Of the same card when, besides making an angle of 45° 
with H, as in last example, the horizontal projection of the 
diagonal makes an angle of 30° with GL. 

28. Same as example 27, except that the horizontal projec- 
tion of the diagonal makes an angle of 90° with GL. 

29. Of an hexagonal card whose sides are £•" long, when 
one of its long diameters is parallel to V and 1" in front of V, 
one end resting on H. The surface of the card is perpendicu- 
lar to V and makes an angle of 30° with H. 

30. Of same card when its diameter, besides making an 
angle of 30° with H, as in last example, has its horizontal pro- 
jection inclined at an angle of 60° with GL. 

31. Of a pentagonal card whose surface is perpendicular to 
H and makes an angle of 45° with V, one edge being perpen- 
dicular to H and resting against V. The diameter of the cir- 
cumscribed circle is 1-J". 



130 EXAMPLES. 

32. Of same card when, besides making an angle of 45° 
with V, as in last example, it has been revolved through an 
angle of 30°. 

33. Of an octagonal card resting on one of its edges, with 
its surface perpendicular to V and inclined at an angle of 60° 
with H. The diameter of the inscribed circle is 1-|". 

34. Of same card when it is inclined at same angle with H 
as in last example, and the horizontal projection of the edges 
which were perpendicular to V makes an angle of 45° with 
GL. 

35. Of an isosceles triangle situated in a profile plane, the 
base making an angle of 15° with H, its back corner being f" 
above H and J" in front of V, and lower than its front corner. 
The base of triangle is 1-J" and altitude 1". 

36. Of an hexagonal card whose surface is perpendicular 
to both V and H. Its long diameter is 1". Two of its edges 
are perpendicular to H. Centre of card is 1" above H, and 
1^" in front of V. 

37. Of a circular card 3" in diameter, perpendicular to H 
and making an angle of 30° with V, resting on H, back edge 
1" from V. 

38. Of same card after it has been revolved through an 
angle of 45°, keeping the same angle with V as in last exam- 
ple. 

39. Of a circular card 3" in diameter, perpendicular to H 
and making an angle of 45° with V. Solve as explained in 
Art. 61. 

40. Draw plan and elevation of a cube of 1" edge resting 
on H, J" in front of V, with two faces parallel to V. Find 
the true size of the angle which the diagonal, which slopes 
downward, backward, and to the right, makes with V and also 
with H. 



EXAMPLES. 131 

41.* Of a rectangular prism whose base is f" x 1" and 
length is 1 J", resting with its base against V, its lower left-hand 
face making an angle of 60° with V. 

Shade lines are to be put in in all the examples where there are any. 

42. Of a cylinder resting with its base on H ; diameter of 
cylinder is 1" and its length is 1|-". The axis is f" in front 
of V. 

43. Of a cone resting with its base parallel to and J" in 
front of V. The diameter of base is 1J" and its height is If". 

44. Of a heptagonal prism resting with its base on H, one 
of its faces making an angle of 15° with V. The diameter of 
the circumscribed circle about base is 1" and the height of 
prism is If". 

45. Of an octagonal pyramid resting with its base on H, 
with two of the edges of the base making an angle of 30° with 
GL. The diameter of the circle inscribed in the base is 1-J" 
and the altitude is 1". 

46. Draw plan and two elevations of a square prism with 
its axis parallel to both V and H ; the axis is 1" above H and 
1J" in front of V, base of prism is f" square, and the length is 
H" ; the upper left-hand long face makes an angle of 30° 
with H. 

47. Draw plan and elevation of same prism when its axis 
is parallel to H and makes an angle of 45° with V, backward 
and to the left. 

48. Draw plan and two elevations of an hexagonal prism 
2" long, diameter of circumscribed circle about base is 1J-", 
the axis is parallel to H and makes an angle of 30° with V, 
backward and to the right ; lower edge of prism rests on H, 
and the lower right-hand face makes an angle of 20° with H. 

* The base is supposed to be at right angles to the axis in all the prisms, pyra- 
mids, cylinders, and cones, unless otherwise stated. 



132 EXAMPLES. 

49. Draw plan and two elevations of a circular cylinder, 
3" in diameter and 2J" long, with a circular hole through it 
1J" in diameter ; axis is parallel to both V and H. 

50. Draw plan and elevation of same cylinder resting with 
its base on H. 

51. Of same cylinder when lying on H, with its axis par- 
allel to H and making an angle of 60° with V. 

52. Draw plan and two elevations of a pile of blocks 
located as follows : the lowest one is 3" long, 1J" wide, and -J-" 
thick, it rests with its wide face on H, its long edge making 
an angle of 30° with V, backward and to the right ; on top of 
this a second block rests, equal in width and thickness to the 
first but 1" shorter ; on this a third block rests, of the same 
width and thickness as the others, but 1" shorter than the sec- 
ond ; these blocks are placed symmetrically. 

53. Draw plan and elevation of a pyramid formed of four 
equilateral triangles of 2" sides, when one edge of the base is 
at 30° with V. 

54. Of an hexagonal prism standing with its base on H, 
two of its faces making angles of 20° with V; diameter of 
circumscribed circle about base is 8", length of prism is 10", 
scale 3" = 1', or £ size. 

55. Of same prism when axis is parallel to V and makes 
an angle of 60° with H and slopes downward to the left. 
Scale 3" = r. 

56. Of same prism when its axis, besides making an angle 
of 60° with H, has its horizontal projection inclined at an angle 
of 60° with GL ; axis of prism slopes downward, forward, and 
to the left. Scale 2"=1'. 

57. Of a pentagonal pyramid resting with its base on H, 
one edge of base perpendicular to V, diameter of circumscribed 
circle 16", height of pyramid 20". Find its shadow. Scale 
l£" = r, or ^ size. 



EXAMPLES. 133 

5&. Of same pyramid when its axis is parallel to V and 
slopes downward to the left, making an angle of 75° with H. 
Find its shadow. Scale 1-J-"=1'. 

59. Of same pyramid when its axis, besides making an angle 
of 75° with H, has its horizontal projection inclined at an angle 
of 30° with GL, so that axis slopes downward, backward, and 
to the right. Find its shadow. Scale 1|-" = 1'. 

60. Of a cone resting with its base on H, diameter, of base 
20", height of cone 2'. Find its shadow. Scale 1-J-" = 1'. 

61. Of same cone when resting with an element on H, 
with its axis parallel to V. Find its shadow. Scale 1J"=1\ 

62. Of same cone with an element on H, and axis making 
an angle of 45° with GL. Find its shadow. Scale 1£" = 1'. 

63. Of same cone with an element on H, and axis lying 
in a profile plane, sloping downward and backward. Find its 
shadow. Scale 1J"=1'. 

64. Of the frustum of an octagonal pyramid resting with 
its base on H, long diameter of lower base is 3" and of the 
upper base is 2", the height of frustum is 3", the front left-hand 
edge of base makes an angle of 15° with GL, backward to the 
left. There is a hole 1" square through the centre of frustum, 
whose axis is coincident with axis of frustum ; two sides of 
the hole make angles of 7£° with GL. 

65. Of same frustum when its axis is parallel to V and 
makes an angle of 60° with H and slopes downward to the left. 

QQ. Of same frustum when its axis, besides making an angle 
of 60° with H, has its horizontal projection inclined at an angle 
of 30° with GL, and slopes downward, forward, and to the left. 

67. Of the skeleton frame of a box 3' long, 2' wide, and 2' 
high, the joists being all 2^" square. The frame rests on H 
with its long sides parallel to V. Do not show joints in framing. 
Scale 1"=1\ 



134 EXAMPLES. 

68. Of same frame still resting on H with its long sides 
making an angle of 50° with V, backward to the left. Scale 
1"=1\ 

69. Revolve the elevation obtained in Example 68 through 
an angle of 30° (in either direction), and construct correspond- 
ing plan. Scale 1"=1'. 

70. Of a double cross standing on its base, one arm paral- 
lel to both V and H; upright piece is 1—8" square and 10'— 8" 
high, each arm is 1—8" square and 7' long (out to out), their 
top surfaces are 2-8" below the top of upright. Scale f"=l'. 

71. Of same cross when its axis is parallel to V and makes 
an angle of 60° with H. Scale |" = 1'. 

72. Of same cross when its axis, besides making an angle 
of 60° with H, has its horizontal projection inclined at an 
angle of 60° with H, sloping downward, forward, and to the 
left. Scale -|" = 1'. 

73. Of a pyramid resting on its apex with axis perpendicu- 
lar to H and 2-6" in front of V, its base is an equilateral tri- 
angle of 2'-8" sides and its altitude is 11". The left-hand edge 
of base is perpendicular to V. Find its shadow. Scale f"=l'. 

74. Of same pyramid when its axis is parallel to V and 
makes an angle of 60° with H, and slopes downward to the 
right. Find its shadow. Scale f"=l'. 

75. Of same pyramid when its axis, besides making an angle 
of 60° with H, has its horizontal projection inclined at an angle 
of 30° with GL, so that it slopes downward, forward, and to 
the left, its apex being l'-6" from V, still resting on H. Find 
its shadow. Scale f"=l'. 

76. There is a solid formed of two equal square pyramids 
of 2" base and 3" altitude, which are united by their bases. 
Draw plan and elevation when the object rests with one of its 
triangular faces on H, its axis being parallel to V and 2£" in 
front of V. Find its shadow. 



EXAMPLES. 135 

77. Of same object still resting on one of its faces, when 
the horizontal projection of the axis makes an angle of 45° 
with GL, and slopes downward, backward, and to the right. 
Find its shadow. 

78. Draw projections of a pentagonal prism whose length 
is 2^-" and radius of circumscribed circle about the end is J" ; 
the prism rests with one of its long edges on H, which makes 
an angle of 60° with V, backward to the left, and whose front 
end is 3" from V. The lower left-hand long face makes an 
angle of 15° with H. 

Also, of a triangular pyramid resting on its base on H, with 
its axis If" to the left of the point located in prism, and 4" in 
front of V, diameter of circumscribed circle is 2", the altitude 
of pyramid is 3^", right-hand edge of base is perpendicular to 
V. Find shadow of prism on H and V, also of pyramid on H 
and on prism. 

79. Find the shadow on H of a card •§" square, parallel to 
H, and •£-" above H, two edges making angles of 30° with V. 

80. Of same card on H, when it is parallel to V, 2" in front 
of V, two edges parallel to H, lowest edge J" above H. 

81. Of same card on V and H, lying in a profile plane, two 
edges perpendicular to H, back edge ^" in front of V and low- 
est edge J" above H. 

82. Of an hexagonal card, parallel to V, %" in front of V, 
two edges parallel to H, centre of hexagon 1J" above H, long 
diameter 1". 

83. Of same card parallel to H, 1-J'. above H, centre 1" in 
front of V, two edges parallel to V. 

84. Of a circular card, 1J" in diameter, parallel to V, 1^" 
in front of V, centre 1" above H. 

85. Of an hexagonal card whose surface is perpendicular 
to V and H, two of its edges perpendicular to H, centre of 



136 EXAMPLES. 

hexagon \%" above H, 1-^" in front of V, diameter of inscribed 
circle 1". In constructing projections of hexagon revolve it 
about a vertical axis through centre. 

86. Of a cube of f" sides, parallel to V and H, 1^" above 
H, and J" in front of V. 

87. Of a square prism standing on H, each face § " x 1^", 
two faces parallel to V, ^" in front of V. 

88. Of same prism still standing on H, turned so that two 
faces make angles of 30° with V, backward to the right. 

89. Of a cylinder f" in diameter and If" high, with base 
resting against VI" above H. 

90. Of a cone 1J" high, base f " in diameter, standing on H, 
with axis If" in front of V. 

91. Of a line located as shown in Example 20. 

92. Of a card located as shown in Example 36. 




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